Question

The electric field strength depends only on the x, y and z coordinates according to the law $E = \frac{a(x\hat{i} + y\hat{j} + z\hat{k})}{(x^2 + y^2 + z^2)^{3/2}}$, where a = 122.5 SI unit and is a constant. Find the potential difference (in volt) between (3, 2, 6) and (0, 3, 4).

Answer 1

7

Answer 2

The electric field is given by $E = \frac{a(x\hat{i} + y\hat{j} + z\hat{k})}{(x^2 + y^2 + z^2)^{3/2}}$.

This can be written as $E = \frac{a\vec{r}}{r^3}$, where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $r = \sqrt{x^2 + y^2 + z^2}$.

This electric field is conservative and can be expressed as the negative gradient of a scalar potential function, $E = -\nabla V$.

The potential function corresponding to this electric field is $V(\vec{r}) = \frac{a}{r} = \frac{a}{\sqrt{x^2 + y^2 + z^2}}$.

We need to find the potential difference between point A (3, 2, 6) and point B (0, 3, 4). The potential difference between A and B is $V_B - V_A$.

First, calculate the distance of point A from the origin:

$r_A = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.

The potential at point A is $V_A = V(3, 2, 6) = \frac{a}{r_A} = \frac{a}{7}$.

Next, calculate the distance of point B from the origin:

$r_B = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5$.

The potential at point B is $V_B = V(0, 3, 4) = \frac{a}{r_B} = \frac{a}{5}$.

The potential difference between (3, 2, 6) and (0, 3, 4) is $V_B - V_A$.

$V_B - V_A = \frac{a}{5} - \frac{a}{7} = a \left(\frac{1}{5} - \frac{1}{7}\right) = a \left(\frac{7 - 5}{35}\right) = \frac{2a}{35}$.

Given $a = 122.5$ SI unit.

Substitute the value of $a$ into the expression for the potential difference:

$V_B - V_A = \frac{2 \times 122.5}{35} = \frac{245}{35} = 7$.

The potential difference between (3, 2, 6) and (0, 3, 4) is 7 Volts.