Each side of a square is 0-9 m long- The charges at its corners are -2 -C-3 -C-4 -C and -5 -C-Determine the electric potential at the centre of the square-

[ (A C)^2=(B D)^2=a^2+a^2$A C)=(B D)=√(2 a^2) A C=B D=√(2 a), a=0.9 m A O=B O=C O=D O=A C2 A O=B O=C O=D O=0.9 √(2)2=0.9√(2)Electric potent ...

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Standard XII

Physics

Electric Potential at a Point in Space Due to a Point Charge Q

Each side of ...

Question

Each side of a square is $0.9 m$ long. The charges at its corners are $- 2 μ C, + 3 μ C, - 4 μ C$ and $+ 5 μ C$ .Determine the electric potential at the centre of the square.

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Solution

$\begin{matrix} (A C)^{2} = (B D)^{2} = a^{2} + a^{2} $ A C) = (B D) = \sqrt{2 a^{2}} A C = B D = \sqrt{2 a}, a = 0.9 m A O = B O = C O = D O = \frac{A C}{2} \end{matrix}$
$A O = B O = C O = D O = \frac{0.9 \sqrt{2}}{2} = \frac{0.9}{\sqrt{2}}$
Electric potential at centre
$\begin{matrix} V_{0} = \frac{1}{4 π ε_{0}} [\frac{- 2 \times 10^{- 6}}{A O} + \frac{3 \times 10^{- 6}}{B O} - \frac{4 \times 10^{- 6}}{C O} + \frac{5 \times 10^{- 6}}{D O}] V_{0} = \frac{9 \times 10^{9} \times 10^{- 6}}{\frac{0.9}{\sqrt{2}}} [- 2 + 3 - 4 + 5] V_{0} = \frac{9 \times 10^{3} \times 10 \times \sqrt{2}}{9} [2] V_{0} = 2 \times 1.414 \times 10^{4} V_{0} = 2.828 \times 10^{4} v o l t \end{matrix}$

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Each side of a square is 0.9m long. The charges at its corners are −2μC,+3μC,−4μC and +5μC.Determine the electric potential at the centre of the square.

(AC)2=(BD)2=a2+a2$AC)=(BD)=√2a2AC=BD=√2a,a=0.9mAO=BO=CO=DO=AC2AO=BO=CO=DO=0.9√22=0.9√2Electric potential at centreV0=14πε0[−2×10−6AO+3×10−6BO−4×10−6CO+5×10−6DO]V0=9×109×10−60.9√2[−2+3−4+5]V0=9×103×10×√29[2]V0=2×1.414×104V0=2.828×104volt

Each side of a square is $0.9 m$ long. The charges at its corners are $- 2 μ C, + 3 μ C, - 4 μ C$ and $+ 5 μ C$ .Determine the electric potential at the centre of the square.