Two points A and B are at a distance a and b respectively from an infinite metal plate having charge density as - - - C m -2- Potential difference between A and B isA- -0 b - a B- -2 -0 a - b C- -0 a - b D- -2 -0 b - a

The correct option is A σ/ϵ _0(b a)As, Δ V = E Δ r ∴ V_A V_B = E(a b) Here, E = σ/ϵ _0⇒ V_A V_B=σ/ϵ _0(b a) ...

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Standard XII

Physics

Equipotential Surfaces

Two points A ...

Question

Two points A and B are at a distance a and b respectively from an infinite metal plate having charge density as ‘ $σ$ ’ C $m^{- 2}$ . Potential difference between A and B is

\frac{σ}{ϵ_{0}} (b - a)

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\frac{σ}{ϵ_{0}} (a - b)

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\frac{σ}{2 ϵ_{0}} (b - a)

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\frac{σ}{2 ϵ_{0}} (a - b)

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Solution

The correct option is A $\frac{σ}{ϵ_{0}} (b - a)$
As, $Δ V = - E Δ r ∴ V_{A} - V_{B} = - E (a - b)$
Here, $E = \frac{σ}{ϵ_{0}} \Rightarrow V_{A} - V_{B} = \frac{σ}{ϵ_{0}} (b - a)$

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Two points A and B are at a distance a and b respectively from an infinite metal plate having charge density as ‘σ’ C m−2. Potential difference between A and B is

The correct option is A σϵ0(b−a)As, ΔV=−EΔr∴VA−VB=−E(a−b) Here, E=σϵ0⇒VA−VB=σϵ0(b−a)

Two points A and B are at a distance a and b respectively from an infinite metal plate having charge density as ‘ $σ$ ’ C $m^{- 2}$ . Potential difference between A and B is

The correct option is A $\frac{σ}{ϵ_{0}} (b - a)$
As, $Δ V = - E Δ r ∴ V_{A} - V_{B} = - E (a - b)$
Here, $E = \frac{σ}{ϵ_{0}} \Rightarrow V_{A} - V_{B} = \frac{σ}{ϵ_{0}} (b - a)$