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Question: If three point charges are placed at the three corners of a square of diagonals 2a as shown in figur...

If three point charges are placed at the three corners of a square of diagonals 2a as shown in figure. Then ∫BAE→.dr→\int_{B}^{A}{\overset{\rightarrow}{E}.\overset{\rightarrow}{dr}} will be -

A

q4π∈0a\frac{q}{4\pi \in_{0}a}

B

–q8π∈0a\frac{–q}{8\pi \in_{0}a}

C

–q4π∈0a\frac{–q}{4\pi \in_{0}a}

D

q8π∈0a\frac{q}{8\pi \in_{0}a}

Answer

–q8π∈0a\frac{–q}{8\pi \in_{0}a}

Explanation

Solution

VB – VA = –=

Now VB=K q2a\frac { q } { 2 a } –kqa\frac { \mathrm { kq } } { \mathrm { a } }=–

\=– = −q8πϵ0a- \frac { \mathrm { q } } { 8 \pi \epsilon _ { 0 } \mathrm { a } }