Question

Three identical charges, each with a value of 1.0 × 10⁻⁸ C, are placed at the corners of an equilateral triangle of side 20 cm. Find the electric field and potential at the centre of the triangle. 

Answer 1

Given:
Magnitude of charges, q =  1.0 × 10⁻⁸ C
Side of the triangle, 

l = 20 cm = 0.2 m  

Let \[E_A ,  E_B  \text{ and }   E_C\] be the electric fields at the centre due to the charges at A, B and C, respectively.

The distance of the centre is the same from all the charges. So,

\[E_A  =  E_B  =  E_C  = E  \left( \text{ say }\right)\]

Resolving \[E_B   \text{ and }   E_C\]  into vertical and horizontal components (with θ = 30° )

 

The horizontal components cancel each other, as shown in the figure.

So, the net electric field at the centre,

\[E_{net}  =  E_A  - \left( E_B \sin\theta + E_A \sin\theta \right)\] 

\[             = E\left( 1 - \sin30^\circ- \sin30^\circ \right)\] 

\[ \Rightarrow  E_{net}  = 0\]

Now,

\[h^2  =  l^2  -  \left( \frac{l}{2} \right)^2 \] 

\[ h^2  =  \left( 0 . 2 \right)^2  -  \left( 0 . 1 \right)^2 \] 

\[ \Rightarrow h = \frac{\sqrt{3}}{10}\]

Let the distance of the centre from each charge be r.
For an equilateral triangle,

\[r = \frac{2}{3}h\] 

\[ \Rightarrow r = \frac{2}{3} \times \frac{1 . 732}{10} = 1 . 15 \times  {10}^{- 1} \]  m

Potential at the centre,

\[V =  V_A  +  V_B  +  V_C \] 

\[ \because    V_A  =  V_B  =  V_{C,} \] 

\[  V = 3 V_A\]

\[V = 3 \times \frac{1}{4\pi \epsilon_0}\frac{q}{r}\] 

\[V = \frac{3 \times 9 \times {10}^9 \times {10}^{- 8}}{0 . 115}\]

\[V = 23 \times  {10}^2  = 2 . 3 \times  {10}^3 \] V