Q.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) $10\ \text{N/C}$ along $+x$ (B) $10\ \text{N/C}$ along $-x$ (C) $50\ \text{N/C}$ along $+x$ (D) $50\ \text{N/C}$ along $-x$

The electric field is the negative gradient of potential; differentiating $V = 10 - 50x$ gives $E = 50\ \text{N/C}$ along the $+x$ direction.

Potential falls with distance from a positive charge, and P is nearer the charge than Q.

For an isolated positive point charge, $V=\dfrac{kq}{r}$, so potential decreases as distance from the charge increases. From the field-line diagram, point P lies closer to the charge than point Q, so $V_P>V_Q$, i.e. $V_P-V_Q>0$.

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:

$E = -\frac{dV}{dr}$

If the field intensity $E = 0$ in a region, then $\frac{dV}{dr} = 0$. 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 $r$.

This is exactly the situation inside a hollow charged conductor or between the plates outside a shielded region: no field, uniform potential.

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:

$E = -\frac{dV}{dr}$

The negative sign shows the field points in the direction of decreasing potential. The magnitude gives the potential drop per unit distance (V/m).

Potential falls in the direction of E, because $E = -\dfrac{dV}{dr}$.

The electric field points in the direction of decreasing potential. This follows from the relation:

$E = -\frac{dV}{dr}$

Since E is taken as positive along the direction of the field, $\dfrac{dV}{dr}$ must be negative along that direction, meaning the potential decreases as you move along the direction of the electric field (and increases if you move against it).

The electric field is the negative gradient of the potential. For $V=-5x+3y+\sqrt{15}z$, each component of $\vec E$ is just minus the coefficient of the corresponding coordinate.

Why: $\vec E=-\nabla V=-\left(\dfrac{\partial V}{\partial x}\hat i+\dfrac{\partial V}{\partial y}\hat j+\dfrac{\partial V}{\partial z}\hat k\right)$. Since $V$ is linear in $x,y,z$, each partial derivative is just the coefficient of that variable, so $\vec E$ is uniform (constant everywhere).

Steps:

1. $E_x=-\dfrac{\partial V}{\partial x}=-(-5)=5$

2. $E_y=-\dfrac{\partial V}{\partial y}=-(3)=-3$

3. $E_z=-\dfrac{\partial V}{\partial z}=-(\sqrt{15})=-\sqrt{15}$

4. Magnitude: $|\vec E|=\sqrt{E_x^2+E_y^2+E_z^2}=\sqrt{5^2+(-3)^2+(-\sqrt{15})^2}=\sqrt{25+9+15}=\sqrt{49}=7$

Potential gradient = dV/dx (magnitude of the electric field).

The potential gradient is the rate at which electric potential changes with position:

$\text{potential gradient} = \frac{dV}{dx}.$

It is related to the electric field by $E = -\dfrac{dV}{dx}$; the field points in the direction of decreasing potential. Its SI unit is volt per metre (V/m). In a potentiometer, the potential gradient (fall of potential per unit length of wire) is a key working quantity.