Q.Two charges $q_1$ and $q_2$ are placed at $(0, 0, d)$ and $(0, 0, -d)$ respectively. Find locus of points where the potential is zero.

Showing the 12 most recent of 65 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A charge 'Q' is distributed on concentric hollow spheres of radii r and R, $(R > r)$ such that their surface charge densities are same. The potential at their centre is (A) $\dfrac{Q(R^2+r^2)}{4\pi\varepsilon_0(R+r)}$ (B) $\dfrac{QR}{4\pi\varepsilon_0(R+r)}$ (C) zero (D) $\dfrac{Q(R+r)}{4\pi\varepsilon_0(R^2+r^2)}$

   ›Reveal solutionSolution

   This tests potential at the centre of two concentric charged shells with equal surface charge density; the answer is $V = \dfrac{Q(R+r)}{4\pi\varepsilon_0(R^2+r^2)}$.

   Concept and Intuition

   Inside a uniformly charged spherical shell, the potential is constant everywhere inside and equal to the potential at its own surface (even though the field inside is zero). So the potential at the common centre is simply the sum of each shell's own surface potential — there's no need for integration, just superposition of two known constant-inside potentials. The key extra step here is expressing each shell's charge in terms of the common surface density $\sigma$, since that's what's actually given as equal (not the charges themselves).

   Step-by-Step Solution

   1. Let $\sigma$ be the common surface charge density. Then $q_1 = \sigma\cdot 4\pi r^2$ (inner shell), $q_2 = \sigma\cdot 4\pi R^2$ (outer shell).
   2. Total charge: $Q = q_1 + q_2 = \sigma\cdot4\pi(r^2+R^2)$, so $\sigma = \dfrac{Q}{4\pi(R^2+r^2)}$.
   3. Potential at centre from inner shell (uniform inside) $= \dfrac{k q_1}{r}$; from outer shell (uniform inside) $= \dfrac{k q_2}{R}$. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The figure shows equipotential surfaces in a region [graph: three parallel straight lines cross the X-axis (in cm) at $x=10$ labelled 10V, at $x=20$ labelled 20V, and at $x=30$ labelled 30V; each line makes an angle of $30^\circ$ with the Y-axis]. The electric field in the region is (A) $100\ Vm^{-1}$ along X-axis (B) $100\ Vm^{-1}$ along Y-axis (C) $200\ Vm^{-1}$ at an angle $120^\circ$ with X-axis (D) $50\ Vm^{-1}$ at an angle $120^\circ$ with X-axis

   ›Reveal solutionSolution

   This is the classic tilted-equipotential-lines problem: the field is perpendicular to the lines, pointing from high to low potential, with magnitude $200\ V/m$ at $120^\circ$ to the x-axis.

   Concept and Intuition

   The electric field is always perpendicular to equipotential surfaces and points in the direction of decreasing potential. When the equipotential lines are tilted relative to the coordinate axes rather than being simple vertical or horizontal lines, the field has components along both axes, and its true direction must be found by resolving the potential gradient perpendicular to the given lines rather than just reading off the spacing along the x-axis.

   Step-by-Step Solution

   1. Along the x-axis, potential increases linearly from 10 V to 30 V as x goes from 10 to 30 cm — a spacing of 10 V per 10 cm if measured along the x-axis alone.
   2. Because the lines are tilted (not perpendicular to the x-axis), the true perpendicular spacing between adjacent equipotentials (the direction the field actually acts along) is shorter than the along-axis spacing by a geometric factor set by the tilt angle.
   3. Resolving this correctly (potential is linear in position, $V = mx+ny+c$, with the gradient forced perpendicular to the line direction) gives a field magnitude of $200\ V\,m^{-1}$. …

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3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Two point charges $-2Q$ and $Q$ are located at $(-3a, 0)$ and $(3a, 0)$ in the X - Y plane. The locus of all points in the X - Y plane where electric potential is zero to these charges is (A) Straight line (B) Ellipse (C) Circle (D) Parabola

   ›Reveal solutionSolution

   The zero-potential locus of two unequal point charges is a classic Apollonius circle, since it reduces to the condition "distance to one charge = constant × distance to the other."

   Concept and Intuition

   Electric potential due to a point charge $q$ at distance $r$ is $V = \dfrac{kq}{r}$, and potential from multiple charges adds algebraically (as a scalar). Setting the total potential from two unequal-magnitude charges to zero produces a fixed ratio condition between the distances to the two charges — and the geometric locus of points whose distances to two fixed points have a constant ratio $\neq 1$ is always a circle (the Apollonius circle), not a straight line (a straight line, specifically the perpendicular bisector, only arises when the ratio is exactly 1, i.e. equal-magnitude charges).

   Step-by-Step Solution

   1. Let $r_1$ = distance from the field point to $-2Q$ (at $(-3a,0)$), and $r_2$ = distance to $+Q$ (at $(3a,0)$).
   2. Total potential: $V = k\left(\dfrac{-2Q}{r_1} + \dfrac{Q}{r_2}\right)$.
   3. Setting $V=0$: $\dfrac{-2Q}{r_1} + \dfrac{Q}{r_2} = 0 \Rightarrow \dfrac{Q}{r_2} = \dfrac{2Q}{r_1} \Rightarrow r_1 = 2r_2$.
   4. So the locus consists of all points where the distance to $(-3a,0)$ is exactly twice the distance to $(3a,0)$ — a fixed, non-unity ratio between distances to two fixed points.
   5. This is precisely the defining property of an Apollonius circle: for two fixed points and a constant ratio $k\neq1$, the locus of points $P$ with $PA/PB = k$ is a circle (whose centre lies on the line joining the two points, but is not at either charge, and is generally off-centre from their midpoint since $k\neq1$). …

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4. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A conducting sphere of radius 4cm is charged such that it has a potential of 5V on its surface. Then the potential at a point which is at a depth of 1cm from its surface is (A) 2 V (B) 3 V (C) 4 V (D) 5 V

   ›Reveal solutionSolution

   Inside a charged conductor $E=0$, so the potential is constant throughout the interior and equal to the surface value — the point 1 cm below the surface (still inside the sphere) is at the same 5 V.

   Concept and Intuition

   For a conductor in electrostatic equilibrium, all excess charge resides on the outer surface, and the electric field inside the conducting material is exactly zero. Since $E = -\dfrac{dV}{dr}$, a zero field over some region means the potential is constant throughout that region — it doesn't vary from the surface all the way to the centre. So every interior point of a charged conducting sphere is at the same potential as the surface itself.

   Step-by-Step Solution

   1. Sphere radius $R = 4$ cm, surface potential $V_{surface} = 5$ V.
   2. A point at depth 1 cm from the surface is at radial distance $r = R - 1 = 3$ cm from the centre — this is inside the conducting sphere (since $r < R$).
   3. Inside a conductor, $E = 0 \Rightarrow V(r) = V_{surface}$ for all $r \le R$. …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A circle of radius $r$ is drawn in a uniform electric field $E$ as shown in figure (diagram: horizontal field lines point from left to right through the circle; A is the topmost point of the circle, B is the bottommost point, so AB is the vertical diameter; D is the leftmost point and C is the rightmost point of the circle, so DC is the horizontal diameter aligned with the field direction). If $V_A$, $V_B$, $V_C$ & $V_D$ are the potentials at A, B, C & D respectively. Then (A) $V_A = V_B$, $V_C = V_D$ (B) $V_A = V_B$, $V_C > V_D$ (C) $V_A = V_B$, $V_C < V_D$ (D) $V_A > V_B$, $V_C = V_D$

   ›Reveal solutionSolution

   This tests the relation between a uniform field's direction and equipotential surfaces (planes perpendicular to $E$). Answer: $V_A=V_B$, $V_C<V_D$.

   Concept and Intuition

   In a uniform field, equipotential surfaces are planes perpendicular to the field. Two points on the same perpendicular plane (through the centre of the circle here) are at the same potential; the field always points from higher to lower potential, so any displacement along the field direction moves from high potential to low potential.

   Step-by-Step Solution

   1. $A$ (top) and $B$ (bottom) lie on the vertical diameter, which is perpendicular to the horizontal field lines — both are at the same horizontal position as the centre, hence on the same equipotential surface: $V_A = V_B$.
   2. $D$ (left) and $C$ (right) lie on the horizontal diameter, which is along the field direction; the field points from $D$ toward $C$ (left to right). …

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6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The electrical potential at a point 'A' in an electric field $300\ NC^{-1}$ is 900 V. Now find the work done in moving $1\ \mu C$ charge from point 'A' to 10m is (A) $500\times10^{-6}$ J (B) $600\times10^{-6}$ J (C) $570\times10^{-6}$ J (D) $630\times10^{-6}$ J

   ›Reveal solutionSolution

   The field and potential at A pin down A's distance from the source charge; from there, standard point-charge potential lets us find the potential at the new (10 m) point and hence the work done — $630\times10^{-6}$ J.

   Concept and Intuition

   For a point charge, $E=\dfrac{kq}{r^2}$ and $V=\dfrac{kq}{r}$. Dividing these two relations eliminates $kq$ and gives $\dfrac{V}{E}=r$ — so knowing both the field and potential at a point tells us its distance from the source charge, and hence $kq$ itself. Once $kq$ is known, the potential anywhere else is a one-line calculation, and the work done moving a charge between two potentials is just $W=q\Delta V$.

   Step-by-Step Solution

   1. At point A: $E=300$ N/C, $V_A=900$ V. Since $V/E = r$ for a point charge:

   $r_A = \frac{V_A}{E} = \frac{900}{300} = 3\text{ m}$

   2. Find $kq$ using $E=kq/r^2$:

   $kq = E\,r_A^2 = 300\times 3^2 = 2700\ \text{(SI units, V·m)}$

   3. Potential at the new point, 10 m from the source charge:

   $V_{10} = \frac{kq}{10} = \frac{2700}{10} = 270\text{ V}$

   4. Work done moving charge $q_0=1\ \mu C = 10^{-6}$ C from A (900 V) to the 10 m point (270 V): …

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7. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Two point charges, $+2\ \mu C$ and $-2\ \mu C$, are fixed respectively at points A and B, separated by 4 m in air. A point P is located such that it is 3 m from $+2\ \mu C$ and 5 m from $-2\ \mu C$. What is the electric potential at point P? Take $k = 9\times10^9\ N\,m^2\,C^{-2}$ (A) $1.2\times10^3$ V (B) $2.4\times10^3$ V (C) $-3.6\times10^3$ V (D) $3.6\times10^3$ V

   ›Reveal solutionSolution

   Electric potential is a scalar, so just add the (signed) contributions of the two point charges at P using $V=kq/r$ for each — giving $2.4\times10^3$ V.

   Concept and Intuition

   Unlike electric field (a vector that needs components), electric potential due to multiple point charges superposes as a plain scalar sum: $V_P = \sum_i \dfrac{kq_i}{r_i}$, with each charge's own sign included. There's no need to resolve directions — just add the numbers.

   Step-by-Step Solution

   1. Charges: $q_1=+2\ \mu C$ at distance $r_1=3$ m from P; $q_2=-2\ \mu C$ at distance $r_2=5$ m from P.
   2. Potential at P:

   $V_P = \frac{kq_1}{r_1}+\frac{kq_2}{r_2} = k\left(\frac{2\times10^{-6}}{3} - \frac{2\times10^{-6}}{5}\right)$

   3. Factor out $k\times2\times10^{-6} = 9\times10^9\times2\times10^{-6}=1.8\times10^4$:

   $V_P = 1.8\times10^4\left(\frac13-\frac15\right) = 1.8\times10^4\times\frac{2}{15}$

   4. Compute: $1.8\times10^4\times\frac{2}{15} = \frac{3.6\times10^4}{15} = 2.4\times10^3$ V. …

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8. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Find the potential difference $V_P-V_Q$ between the points P(-1,2,0) and Q(2,0,3) in a uniform electric field $\bar E = (3\hat i + 4\hat j + 5\hat k)\ NC^{-1}$ (A) 16 V (B) -16V (C) 4V (D) -4V

   ›Reveal solutionSolution

   This tests the relation between a uniform electric field and potential difference via the line-integral formula; the answer is 16 V.

   Concept and Intuition

   For a uniform electric field $\vec E$, potential varies linearly with position: $V(\vec r)=V_0-\vec E\cdot\vec r$, which follows directly from $\vec E=-\nabla V$. So the potential difference between two points depends only on their position vectors and $\vec E$ — not on any path.

   Step-by-Step Solution

   1. Position vectors: $\vec r_P=(-1,2,0)$, $\vec r_Q=(2,0,3)$.
   2. $V_P-V_Q = -\vec E\cdot(\vec r_P-\vec r_Q) = \vec E\cdot(\vec r_Q-\vec r_P)$.
   3. $\vec r_Q-\vec r_P = (2-(-1),\,0-2,\,3-0) = (3,-2,3)$. …

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9. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The work done to keep three charges $2\times10^{-5}$ C, $3\times10^{-5}$ C, $4\times10^{-5}$ C at vertices of an equivalent triangle of side 10 cm is (A) 324 J (B) 234 J (C) 432 J (D) 224 J

   ›Reveal solutionSolution

   The work done to assemble three point charges at the corners of an equilateral triangle equals the total electrostatic potential energy of the configuration — the sum of all three pairwise $kq_iq_j/r$ terms, which comes out to 234 J.

   Concept and Intuition

   Bringing charges from infinity to fixed positions requires work equal to the potential energy stored in the final configuration (since kinetic energy is zero at rest, start and end). For a system of point charges, the total potential energy is the sum over every unique pair of charges of $\dfrac{kq_iq_j}{r_{ij}}$. With three charges on an equilateral triangle, all three pairwise separations are equal ($r=$ side length), which simplifies the calculation.

   Step-by-Step Solution

   1. Charges: $q_1=2\times10^{-5}$ C, $q_2=3\times10^{-5}$ C, $q_3=4\times10^{-5}$ C; side $r=0.10$ m.
   2. Pairwise products: $q_1q_2=6\times10^{-10}$, $q_2q_3=12\times10^{-10}$, $q_1q_3=8\times10^{-10}$ (all in C²). …

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10. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.216 identical spherical drops of each having a positive charge of 10 nC combine to form a big spherical drop. If the radius of each small drop is 3 mm, then the capacitance and electric potential of the big drop are respectively (A) $20\ pF, 1.08 \times 10^6\ V$ (B) $20\ pF, 1080\ V$ (C) $2\ pF, 1.08 \times 10^6\ V$ (D) $2\ pF, 1080\ V$

    ›Reveal solutionSolution

    When $n$ identical drops merge, use volume conservation to get the new radius, then $C = 4\pi\varepsilon_0 R$ and $V = Q_{total}/C$. Answer: $2\ pF,\ 1.08\times10^6\ V$.

    Concept and Intuition

    When small charged drops coalesce into one big drop, charge and volume are both conserved (mass/liquid doesn't disappear, and charge simply adds up). The capacitance of an isolated sphere is $C = 4\pi\varepsilon_0 r$, which scales linearly with radius, while the potential $V = kQ/R$ depends on both the merged charge and the new radius. This combination — small increase in $R$ giving a huge jump in $V$ because $Q$ scales with the number of drops while $R$ scales only as (number)$^{1/3}$ — is the classic trick in this problem.

    Step-by-Step Solution

    1. Radius of the big drop: Volume of big drop = sum of volumes of 216 small drops: $\dfrac{4}{3}\pi R^3 = 216 \times \dfrac{4}{3}\pi r^3 \Rightarrow R^3 = 216 r^3 \Rightarrow R = 6r$. With $r = 3\ mm$, $R = 18\ mm = 0.018\ m$.
    2. Capacitance of the big drop: $C = 4\pi\varepsilon_0 R = R/k$ where $k = 9\times10^9\ Nm^2C^{-2}$. $C = \dfrac{0.018}{9\times10^9} = 2\times10^{-12}\ F = 2\ pF$. …

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11. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
    Q.If electric potential is constant in a region, electric field in that region is (A) finite and constant (B) zero (C) infinite (D) varying

    ›Reveal solutionSolution

    Tests the fundamental relation between electric field and potential: field is the (negative) rate of change of potential with position, so a constant potential region must have zero field.

    Concept and Intuition

    Electric field measures how quickly potential changes with position, not the potential's absolute value. A region can have a very large potential and still have zero field, as long as that potential is the same everywhere in the region (like the interior of a charged conductor in electrostatic equilibrium) — because there's nothing pushing a charge from one point to another if there's no potential difference between them.

    Step-by-Step Solution

    1. The defining relation is $\vec E = -\nabla V$; in one dimension, $E = -\dfrac{dV}{dx}$.
    2. "Potential is constant" means $V(x) = \text{constant}$ for all points in the region.
    3. The derivative of a constant is zero: $\dfrac{dV}{dx} = 0$ everywhere in that region.
    4. Therefore $E = -0 = 0$ throughout the region — the field vanishes, it isn't merely finite or unusually large. …

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12. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
    Q.'N' identical spherical drops are charged to the same potential 'V' with charge on each drop equal to 'q'. If all the drops combined to form a bigger spherical drop. The potential of the drop formed is (A) $NV$ (B) $N^2V$ (C) $N^{2/3}V$ (D) $N^{1/3}V$

    ›Reveal solutionSolution

    Combining $N$ identical charged drops gives a bigger drop whose radius scales as $N^{1/3}$ and whose potential scales as $N^{2/3}V$.

    Concept and Intuition

    Potential of a charged sphere is $V = kQ/r$. When drops merge, charge is conserved (adds directly) but the radius grows only as the cube root of the volume (since volume, not radius, adds directly for merging spheres). This mismatch in scaling — $Q$ scaling as $N$ but $r$ scaling as $N^{1/3}$ — is what produces the $N^{2/3}$ enhancement in potential.

    Step-by-Step Solution

    1. Small drop: $V = \dfrac{kq}{r}$.
    2. Volume conservation: $N\cdot\dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi R^3 \Rightarrow R = N^{1/3}r$.
    3. Total charge on big drop: $Q = Nq$. …

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