Question

4 indenticalcharges each of value 2 mc are placed at the four corners of square 2 meters . what is the electric potential at the centre of the square​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Electric\:potential\:at\:centre=3.6\sqrt{2}\times 10^{4}\:J/C}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies 4 \: identical \: charges = 2 \mu \: C \: \: \: \: \: \: (Each) \\ \\ \tt: \implies Side \: of \: square = 2 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Electric \: potential \: at \: centre( V_{c}) = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies Length \: of \: diagonal(AC)= \sqrt{2} a \\ \\ \tt: \implies Length \: of \: diagonal(AC) = \sqrt{2} a \\ \\ \tt: \implies Length \: of \: half \: diagonal(AO)= \frac{ 2\sqrt{2} }{2} \\ \\ \tt: \implies Length \:of \: half \: diagonal(AO)= \sqrt{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies V_{c} = V_{1} + V_{2} + V_{3} + V_{4} \\ \\ \tt \because \: V_{1} = V_{ 2} = V_{3} = V_{4} \\ \\ \tt: \implies V_{c} = 4 V_{1} \\ \\ \tt: \implies V_{c} = 4 \times \frac{k q_{1}}{ r_{1}} \\ \\ \tt: \implies V_{c} = 4 \times \frac{9 \times {10}^{9} \times 2 \times {10}^{ - 6} }{\sqrt{2}} \\ \\ \tt: \implies V_{c} =4 \times 9 \times \sqrt{2} \times {10}^{9 - 6} \\ \\ \tt: \implies V_{c} =36\sqrt{2} \times {10}^{3} \\ \\ \green{\tt: \implies V_{c} =3.6\sqrt{2} \times {10}^{4} \: J /c} \\ \\ \green{\tt \therefore Electric \: potential \: at \: center \: is \: 3.6\sqrt{2} \times {10}^{4} \: J/C}$

Answer 2

★ Refer to the attachment for diagram

Given ,

4 indentical charges each of value 2 mc are placed at the four corners of square of side 2 meters

Thus , half of diagonal (r) = 2✓2/2 = ✓2 m

Let , the net electric potential at the center of the square be " v "

We know that that , the electric potential due to point charge at point p is given by

Potential (v)

$\boxed{ \sf{electric \: potential \: (v) = \frac{kq}{r} }}$

Electric potential is scalar quantity

Thus ,

$\tt \mapsto v =4 \times 9 \times {(10)}^{9} \bigg \{ \frac{2 \times {(10)}^{ - 3} }{ \sqrt{2} } \bigg \}$

$\tt \mapsto v = 36\times {(10)}^{6} \times \frac{2}{ \sqrt{2} }$

$\tt \mapsto v = \frac{72 \sqrt{2} \times {(10)}^{ 6} }{2}$

$\tt \mapsto v = 36 \sqrt{2} \times {(10)}^{6}$

$\tt \mapsto v = 50.904 \times {(10)}^{6} \: \: volt$

★ The electric potential at center of square is 50.904 × 10^(6)