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Question

Physics Question on Electric charges and fields

An infinite sheet carrying a uniform surface charge density o lies on the xyx y-plane. The work done to carry a charge qq from the point A=a(i^+2j^+3k^)A =a(\hat{ i }+2 \hat{ j }+3 \hat{ k }) to the point B=a(i^2j^+6k^)B =a(\hat{ i }-2 \hat{ j }+6 \hat{ k }) (where aa is a constant with the dimension of length and ε0\varepsilon_{0} is the permittivity of free space) is

A

3σaq2ε0\frac{3\sigma aq}{2\varepsilon_{0}}

B

2σaqε0\frac{2\sigma aq}{\varepsilon_{0}}

C

5σaq2ε0\frac{5\sigma aq}{2\varepsilon_{0}}

D

3σaqε0\frac{3\sigma aq}{\varepsilon_{0}}

Answer

3σaq2ε0\frac{3\sigma aq}{2\varepsilon_{0}}

Explanation

Solution

The given
A=a(i^+2j^+3k^)A =a(\hat{ i }+2 \hat{ j }+ 3 \hat{ k })
B=a(i^2j^+6k^)B =a(\hat{ i }-2 \hat{ j }+6 \hat{ k })
AB=OBOAA B = O B - O A
=a(i^2j^+6k^)a(i^+2j^+3k^)=a(\hat{ i }-2 \hat{ j }+6 \hat{ k })-a(\hat{ i }+2 \hat{ j }+ 3 \hat{ k })
AB=a(4j^+3k^)A B =a(-4 \hat{ j }+3 \hat{ k })
Work done =q(σ2ε0)k^AB=q\left(\frac{\sigma}{2 \varepsilon_{0}}\right) \hat{ k } \cdot A B (along to ZZ-axis)
=q(σ2ε0)k^a(4j^+3k^)=3qσa2ε0=q\left(\frac{\sigma}{2 \varepsilon_{0}}\right) \hat{ k } \cdot a(-4 \hat{ j }+3 \hat{ k })=\frac{3 q \sigma a}{2 \varepsilon_{0}}
(i^i^=j^j^=k^k^=1(\because \hat{ i } \cdot \hat{ i }=\hat{ j } \cdot \hat{ j }=\hat{ k } \cdot \hat{ k }=1 and i^j^=j^k^=k^i^=0)\hat{ i } \cdot \hat{ j }=\hat{ j } \cdot \hat{ k }=\hat{ k } \cdot \hat{ i }=0)