Question

5. Electric charges of +10uC, +5°C, -3uC and +8uC are placed at the corners of a square of side √2 m. The potential
at the centre of the square is
1) 1.8 V
2) 1.8 x 10^6 V
3) 1.8 x 10^5V
4) 1.8 x 10^4 V
​

Answer 1

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 charge :

$\purple{\boxed{V=\frac{k\ Q}{r}}}$

• 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