Question

Electric charges of +10°C, +5°C, -34C and +8uC are placed at the corners of a square of side
V2 m, the potential at the centre of the square is:
a) 1.8 V
b) 1.8 x 106 V
c) 1.8 x 105 V
d) 1.8 104​

Answer 1

sorry I don't know

Explanation:

Again sorry i dont know

Answer 2

Given :-

• Let  ABCD is a square of side √2 m. Four Electric Charges are placed at the corners of the square. Charge +10μC   is  placed at A and

• 5μC at B, -3μ C, and 8μC are placed at D.

To be found :-

• The Potential at the center of the square.

Theory :-

• The electric potential at a point define is the external work done by a unit positive charge from infinity to that point without any acceleration.

• Electric potential due to two-point - :

• $\purple{\boxed{V=\frac{k\ Q}{r}}}$

Solution :-

• Let the center of the square be O.

• Let $\sf Q_1=10\mu\ C$

• $\sf Q_2=5\mu\ C$

• $\sf Q_3=-3\mu\ C$

• $\sf Q_4=8\mu\ C$

From the Diagram :-

In ∆BDC

CB²+CD²=BD² [ Pythagoras theorem ]

• ⇒ BD²= (√2)² + (√2)²

• ⇒ BD² = 4m

• ⇒ BD = √4m

• ⇒ BD = 2m

Hence , OD = OB = AO = OC = 1m

• We have to find the potential in the center of the square.

• $\sf V{o} =V_{1} + V_{2} + V_{3} + V_{4}$

• $\sf V{o}= \frac{k\ Q_1}{OA}+ \frac{k\ Q_2}{OB}+ \frac{k\ Q_3}{OC}+ \frac{k\ Q_4}{OD}$

Now put the given values!

• $\sf V{o} = \frac{k \times 10 \times 10^{-6}}{1}+ \frac{k \times 5 \times 10^{-6}}{1}+\frac{k \times -3 \times 10^{-6}}{1}+\frac{k \times 8 \times 10^{-6}}{1}$

• $\sf V_{o} = k \times 10^{-6}(10 + 5 - 3 + 5)$

• $\sf V_{o} = k \times(15+5) \times 10^{-9}$

• If the medium between the two points is air then,

• $\sf\:k=\frac{1}{4\pi\:E_{o}}=9\times10^{9}Nm^{2}C^{-2}$

• $\sf V_{o} = 9 \times 10^9 \times 20 \times 10^{-9}$

• $\sf V_{o} = 18 \times 10^4 V$

• $\sf V_{o} = 1.8 \times 10^5 V$

• Therefore, the potential at the center of the square is

• $\sf = 1.8 \times 10^5 V$

Hence,

Correct option 3) 1.8 x 10^5 V ✔

Extra Information:-

• 1) SI unit of electric potential is J/c

• 2) Electric potential is a scalar quantity