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CAF123

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## Homework Statement

A charge q is a distance d from the centre of an earthed conducting sphere of radius a

(d > a).

Given that the image charge for this system is is q

_{1}located a distance b from the centre of the sphere where ##q_1 = -\frac{aq}{d},\,\,b = \frac{a^2}{d}##,

calculate an expression for the E field at point P on the surface of the sphere, directly above the centre of the sphere.

## The Attempt at a Solution

So q is a distance ##\sqrt{a^2 + d^2}## from the centre. q

_{1}is a distance ##\sqrt{b^2 + a^2} = \sqrt{(a^4/d^2) + a^2} ##away. Put these into the expression for potential and sum them together gives: $$V_T = V_q + V_{q_1} = \frac{q}{4 \pi \epsilon_o} \left[\frac{1}{\sqrt{a^2 +d^2}} - \frac{1}{\sqrt{a^2 + d^2}}\right] = 0$$

The E field will be perpendicular to the surface equipotential, and so will point in -y direction assuming positive y is defined vertically upwards. I know in general E = -∇V, but the above is zero (as expected). Should I instead just consider the total electric field using the value of b given?

This would give: $$E_T = \frac{q}{4 \pi \epsilon_o} \left[\frac{1}{a^2 +d^2}\underline{r} - \frac{1}{(a^4/d^2) + a^2}\underline{r'}\right] ,$$ where ##\underline{r}## is the direction of the E field vector (compts in both x and y) for q and similarly for ##\underline{r'}## for the E field direction by ##q_1##.

I think my next step would be to write the vector r decomposed into x and y and then see that the x components of the sum of the two E field vectors should cancel, although I am not sure how to do this.

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