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प्रश्न
Two charges q1 and q2 are placed at (0, 0, d) and (0, 0, – d) respectively. Find locus of points where the potential a zero.
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उत्तर
Following the principle of superposition of potentials as
described in last section, let us find the potential V due to a collection of discrete point charges q1, q2, …, qn, at a point P.
| The potential at P due to the system of point charges is given as the sum of their individual potentials at P, V = `1/(4piε_0) sum q_i/r_i` |
As we know, the potential at point P is V = `sumV_i`,
Where `V_i = q_i/(4piε_0) ; r_i` = magnitude of position vector P relative to qr
Then `V = 1/(4piε_0) sum q_i/r_"pi"`
Let us take a point on the required plane as (x, y, z). The two charges lies on z-axis at a separation of 2d. The potential at the point P due to two charge is given by
`q_1/(sqrt(x^2 + y^2 + (z - d)^2)) + q_2/(sqrt(x^2 + y^2 + (z + d)^2))` = 0
∴ `q_1/(sqrt(x^2 + y^2 + (z - d)^2)) = (-q_2)/(sqrt(x^2 + y^2 + (z + d)^2))`
On squaring and simplifying, we get
`x^2 + y^2 + z^2 + [((q_1/q_2)^2 + 1)/((q_1/q_2)^2 - 1)] (2zd) + d^2 - 0`
The standard equation of sphere is `x^2 + y^2 + z^2 + 2ux + 2uy + 2wz + g` = 0
With centre `(-u, -v, -w)` and radius `sqrt(u^2 + v^2 + w^2) - g`
Hence centre of sphere will be `(0, 0 - d[(q_1^2 + q_2^2)/(q_1^2 - q_2^2)])`
And radius is `r = sqrt((d[(q_1^2 + q_2^2)/(q_1^2 - q_2^2)])^2 - d^2) = (2q_1q_2d)/(q_1^2 - q_2^2)`
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