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 given by $V(x, y, z) = ax^2 + ay^2 + abz^2$. Given $b=2$, so $V(x, y, z) = a(x^2 + y^2) + 2az^2$.

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 $W_{A \to O} = 50 \mu J = 50 \times 10^{-6} J$.

The work done by the electric field is $W_{A \to O} = q(V_A - V_O)$.

$V_A = V(0, 0, 0.1) = a(0^2 + 0^2) + 2a(0.1)^2 = 2a(0.01) = 0.02a$.

$V_O = V(0, 0, 0) = a(0^2 + 0^2) + 2a(0)^2 = 0$.

$W_{A \to O} = q(V_A - V_O) = (4 \times 10^{-6})(0.02a - 0) = 0.08a \times 10^{-6}$.

Given $W_{A \to O} = 50 \times 10^{-6} J$, so $0.08a = 50$, which gives $a = \frac{50}{0.08} = \frac{5000}{8} = 625$.

The electric potential is $V(x, y, z) = 625(x^2 + y^2) + 1250z^2$.

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

Substitute $V = 6250$ and $z = \sqrt{2}$ into the potential equation:

$6250 = 625(x^2 + y^2) + 1250(\sqrt{2})^2$.

$6250 = 625(x^2 + y^2) + 1250(2)$.

$6250 = 625(x^2 + y^2) + 2500$.

$625(x^2 + y^2) = 6250 - 2500 = 3750$.

$x^2 + y^2 = \frac{3750}{625} = 6$.

The equation $x^2 + y^2 = 6$ represents a circle in the xy-plane with radius $r = \sqrt{6}$.

We are given that the radius is $\sqrt{\alpha}$ m.

So, $\sqrt{\alpha} = \sqrt{6}$, which implies $\alpha = 6$.