Q.[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 $V_P$ is electric potential at P and $V_Q$ is electric potential at Q then -

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, po …

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 $r$.

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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:

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

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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:

$E = -\frac{dV}{dr}$ …

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

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

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

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

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