Question

Four charges $q _1=2 \times 10^{-8} C , q _2=-2 \times 10^{-8} C , q _3=-3 \times 10^{-8} C$, and $q _4=6 \times 10^{-8} C$ are placed at four corners of a square of side $\sqrt{2} m$. What is the potential at the centre of the square?

• A. $270 V$
• B. $300 V$
• C. Zero
• D. $100 V$

Answer 1

Answer: (A)

$AC = BD =\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2 m$

$\therefore DO = OB = AO = OC =\frac{2}{2}=1 m$
$\therefore$ Potential at the centre $O , V = k \frac{ q }{ r }$
$V=k\left[\frac{2 \times 10^{-8}}{1}+\frac{-2 \times 10^{-8}}{1}+\frac{-3 \times 10^{-8}}{1}+\frac{6 \times 10^{-8}}{1}\right]$
$V = k \times 3 \times 10^{-8}=9 \times 10^9 \times 3 \times 10^{-8}$ volt
$=27 \times 10=270 V$