Question

The electric potential in a region is given by $V(x, y, z) = ax^2 + ay^2 + abz^2$. 'a' is a positive constant of appropriate dimensions and b, a positive constant such that V is in volts when x, y, z are in meter. Let b = 2. The work done by electric field when a point charge +4 µC moves from the point (0, 0, 0.1) to origin is 50 µJ. The radius of the circle of the equipotential curve corresponding to V = 6250 volts and z = $\sqrt{2}$ m is $\sqrt{\alpha}$ m. Find $\alpha$.

Answer 1

6

Answer 2

The electric potential is $V(x, y, z) = ax^2 + ay^2 + abz^2$. With $b=2$, $V(x, y, z) = a(x^2 + y^2 + 2z^2)$.

The work done by the electric field when a charge $q = +4 \, \mu C = 4 \times 10^{-6} \, C$ moves from point $A = (0, 0, 0.1)$ to origin $O = (0, 0, 0)$ is $W_E = 50 \, \mu J = 50 \times 10^{-6} \, J$.

The work done by the electric field is $W_E = - \Delta U = - q (V_O - V_A)$.

Potential at A: $V_A = a(0^2 + 0^2 + 2(0.1)^2) = a(2 \times 0.01) = 0.02a$.

Potential at O: $V_O = a(0^2 + 0^2 + 2(0)^2) = 0$.

$W_E = - (4 \times 10^{-6}) (0 - 0.02a) = (4 \times 10^{-6})(0.02a) = 8 \times 10^{-8} a$.

Given $W_E = 50 \times 10^{-6}$.

$8 \times 10^{-8} a = 50 \times 10^{-6} \implies a = \frac{50 \times 10^{-6}}{8 \times 10^{-8}} = \frac{50}{8} \times 10^2 = \frac{25}{4} \times 100 = 625$.

The potential function is $V(x, y, z) = 625(x^2 + y^2 + 2z^2)$.

We need the radius of the equipotential curve $V = 6250$ volts at $z = \sqrt{2}$ m.

$625(x^2 + y^2 + 2(\sqrt{2})^2) = 6250$.

$625(x^2 + y^2 + 2 \times 2) = 6250$.

$625(x^2 + y^2 + 4) = 6250$.

$x^2 + y^2 + 4 = \frac{6250}{625} = 10$.

$x^2 + y^2 = 6$.

This is the equation of a circle in the xy-plane with radius $r = \sqrt{6}$.

The radius is given as $\sqrt{\alpha}$. So, $\sqrt{\alpha} = \sqrt{6}$, which means $\alpha = 6$.