The absolute potential due to the two charge configuration as shown in figure at a point P is:

Zero

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\frac{1}{4 π ϵ_{0}} . \frac{+ q}{\sqrt{a^{2} + b^{2}}}

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\frac{1}{4 π ϵ_{0}} . \frac{- q}{\sqrt{a^{2} + b^{2}}}

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\frac{2 q}{4 π ϵ_{0} \sqrt{a^{2} - b^{2}}}

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Solution

The correct option is A Zero
$P o t e n t i a l a t p o i n t P = \frac{k (+ q)}{\sqrt{a^{2} + b^{2}}} + \frac{k (- q)}{\sqrt{a^{2} + b^{2}}}$
Potential = 0 V

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Similar questions

Q. Let

P (a sec θ, b tan θ)

and

Q (a sec ϕ, b tan ϕ),

where

θ + ϕ = \frac{π}{2}

, be two points on the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

. If

(h, k)

is the point of intersection of normals at

P

and

Q

, then

k

is equal to

Let $P (a s e c θ, b t a n θ)$ and $Q (a s e c ϕ, b t a n ϕ)$ , where $θ + ϕ = \frac{π}{2}$ , be two points on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .
If (h, k) is the point of the intersection of the normals at P and Q, then k is equal to