Q.The electric potential along the x-axis in a certain region is given by V(x)=100−25x2 (in volts, with x in metres). Find the electric field at x=2 m.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Relation Between Electric Field and Potential
Field and potential, as vector and scalar descriptions of the same electrostatic field, are linked by E=−dV/dl: the field along any direction equals minus the rate at which potential falls off in that direction, so E points toward the steepest DECREASE of V. Run the other way, VB−VA=−∫ABE⋅dl. For a uniform field (as between …
E=−dV/dx; differentiate the given V(x) and substitute x=2. …
Given V(x)=100−25x2, the field along x is E=−dV/dx:
dxdV=−50x⟹E=−(−50x)=50x
At x=2 m: E=50(2)=100 V/m. Since E comes out positive, the field points in the +x direction (potential decreases as x increases, consistent with …
Differentiate V(x) with respect to x, negate, and substit …
- CBSE 2026Set 55/1/11 markMCQQ.In a region, the electric potential varies as V=10−50x, where V is in volts and x in metres. The electric field in the region is (A) 10 N/C along +x (B) 10 N/C along −x (C) 50 N/C along +x (D) 50 N/C along −x
›Reveal solutionSolution
The electric field is the negative gradient of potential; differentiating V=10−50x gives E=50 N/C along the +x direction.
The connection between electric potential and electric field is one of the most fundamental relationships in electrostatics. Potential tells us the energy landscape; the field tells us which way a positive charge would be pushed and how hard.
The electric field is defined as the negative gradient of the potential:
E=−∇V=−dxdVi^−dydVj^−dzdVk^
The negative sign encodes a physical truth: electric field points from high potential to low potential, in the direction a positive charge naturally moves (downhill in energy). When potential decreases in some direction, the field points in that direction.
In this problem the potential varies only with x, so we have a one-dimensional situation.
Finding the electric field:
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Differentiate the potential with respect to x:
Given V=10−50x, we compute
dxdV=dxd(10−50x)=−50 V/m
- Apply the negative sign to get the field:
Ex=−dxdV=−(−50)=50 V/m
Since 1 V/m=1 N/C, we have
Ex=50 N/C …
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- CBSE 2026Set ANNUAL1 markMCQQ.[FIGURE: Electric field lines radiating outward from a positive point charge; point P lies closer to the charge, point Q lies farther away along a field line.] Above figure shows electric field lines due to a positive point charge. If VP is electric potential at P and VQ is electric potential at Q then -(i) VP−VQ>0(ii) VP−VQ<0(iii) VP−VQ=0(iv) [TEXT CUT OFF AT PAGE EDGE in source render — not legible]
›Reveal solutionSolution
Potential falls with distance from a positive charge, and P is nearer the charge than Q.
For an isolated positive point charge, V=rkq, so potential decreases as distance from the charge increases. From the field-line diagram, po …
- CBSE 2026Set A1 markMCQQ.In an electric field with intensity E = 0, the change of potential V with distance r will be (A) V ∝ 1/r^2 (B) V ∝ 1/r (C) V ∝ r (D) V = constant and independent of r
›Reveal solutionSolution
E and V are linked by E = -dV/dr; if E = 0 everywhere then V is constant, independent of r.
The electric field is the negative gradient of potential:
E=−drdV
If the field intensity E=0 in a region, then drdV=0. A quantity whose derivative with respect to r is zero cannot change with r — it must be a constant. So the potential has the same value everywhere in that region regardless of r.
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- CBSE 2024Set A1 markMCQQ.The relation between electric field (E) and electric potential (V) is (A) E = -(dV/dr) (B) E = -(dr/dV) (C) E = (dV/dr) (D) E = (dr/dV)
›Reveal solutionSolution
Electric field equals the negative rate of change of potential with distance: E = −dV/dr.
The electric field is the negative gradient of the electric potential:
E=−drdV
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- CBSE 2024Set A1 markQ.Fill in the blank with appropriate word: Potential ______ on moving along the direction of electric field.
›Reveal solutionSolution
Potential falls in the direction of E, because E=−drdV.
The electric field points in the direction of decreasing potential. This follows from the relation:
E=−drdV …
- CBSE 2023Set ANNUAL1 markMCQQ.Electric potential at any point is V = -5x + 3y + √15z, then the magnitude of electric field is –(a) 3√2(b) 4√2(c) 5√2(d) 5
›Reveal solutionSolution
The electric field is the negative gradient of the potential. For V=−5x+3y+15z, each component of E is just minus the coefficient of the corresponding coordinate.
Why: E=−∇V=−(∂x∂Vi^+∂y∂Vj^+∂z∂Vk^). Since V is linear in x,y,z, each partial derivative is just the coefficient of that variable, so E is uniform (constant everywhere).
Steps:
- Ex=−∂x∂V=−(−5)=5
- Ey=−∂y∂V=−(3)=−3
- Ez=−∂z∂V=−(15)=−15 …
- CBSE 2021Set A1 markMCQQ.Potential gradient is equal to (A) dx/dV (B) dx . dV (C) dV/dx (D) None of these
›Reveal solutionSolution
Potential gradient = dV/dx (magnitude of the electric field).
The potential gradient is the rate at which electric potential changes with position:
potential gradient=dxdV.
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