Question

3. Charges 5 uc, -2 °C, 3 PC and -9uC are placed
at the corners A, B, C and D of a square ABCD of
side 1m. The net electric potential at the centre of
the square is
1)-27 KV
2) -27 13 KV
3) - 90KV
4) zero​

Answer 1

Given:

Charges 5 uc, -2 uC, 3 uC and -9uC are placed

at the corners A, B, C and D of a square ABCD of

side 1m.

To find:

Net electric potential at centre?

Calculation:

• Distance of centre from vertex = d/√2 = 1/√2 metres.

Now, field potential is :

$\rm V_{net} = \bigg \{\dfrac{k(5)}{r} + \dfrac{k( - 2)}{r} + \dfrac{k(3)}{r} +\dfrac{k( - 9)}{r} \bigg \} \times {10}^{ - 6}$

$\rm \implies V_{net} = \dfrac{k(5 - 2 + 3 - 9)}{r} \times {10}^{ - 6}$

$\rm \implies V_{net} = \dfrac{k( - 3)}{r} \times {10}^{ - 6}$

$\rm \implies V_{net} = \dfrac{ - 3 \times (9 \times {10}^{9} )}{ \frac{1}{ \sqrt{2} } } \times {10}^{ - 6}$

$\rm \implies V_{net} = - 27 \sqrt{2} \times {10}^{3} \: volt$

So, net potential is -27√2 kilo-volts.

Answer 2

Answer:

$Given:</p><p></p><p>Charges 5 uc, -2 uC, 3 uC and -9uC are placed</p><p></p><p>at the corners A, B, C and D of a square ABCD of</p><p></p><p>side 1m.</p><p></p><p>To find:</p><p></p><p>Net electric potential at centre?</p><p></p><p>Calculation:</p><p></p><p>Distance of centre from vertex = d/√2 = 1/√2 metres.</p><p></p><p>Now, field potential is :</p><p></p><p>\rm V_{net} = \bigg \{\dfrac{k(5)}{r} + \dfrac{k( - 2)}{r} + \dfrac{k(3)}{r} +\dfrac{k( - 9)}{r} \bigg \} \times {10}^{ - 6}Vnet={rk(5)+rk(−2)+rk(3)+rk(−9)}×10−6</p><p></p><p>\rm \implies V_{net} = \dfrac{k(5 - 2 + 3 - 9)}{r} \times {10}^{ - 6}⟹Vnet=rk(5−2+3−9)×10−6</p><p></p><p>\rm \implies V_{net} = \dfrac{k( - 3)}{r} \times {10}^{ - 6}⟹Vnet=rk(−3)×10−6</p><p></p><p>\rm \implies V_{net} = \dfrac{ - 3 \times (9 \times {10}^{9} )}{ \frac{1}{ \sqrt{2} } } \times {10}^{ - 6}⟹Vnet=21−3×(9×109)×10−6</p><p></p><p>\rm \implies V_{net} = - 27 \sqrt{2} \times {10}^{3} \: volt⟹Vnet=−272×103volt</p><p></p><p>So, net potential is -27√2 kilo-volts.</p><p></p><p>$

solution by rohit