Question

5.
A charge of 2C is placed on the x-axis at Im from the origin along -ve x-axis. Infinite number of charges each
of magnitude 2C are placed on x-axis at lm, 2m, 4m, .... from origin along +ve x-axis. The first charge is
positive and alternate charges are of opposite in nature. The electric field intensity at the origin​

Answer 1

A charge 2C is placed on the x-axis at 1 m from the origin along -ve x-axis. infinite number of charges each of magnitude 2C are placed on x-axis at 1 m, 2 m , 4 m ...... from origin along +ve x-axis. The first charge is positive and alternative charge are opposite in nature.

To find : The electric field intensity at the origin.

solution : electric field due to 1st charge at origin, E₁ = k(2C)/(1m)² [ along +ve x-axis direction ]

electric field due to 2nd charge at origin, E₂ = k(2C)/(2m)² [ along -ve x - axis direction ]

electric field due to 3rd charge at origin, E₃ = k(2C)/(3m) [ along +ve x-axis direction ]

......... ..... so on.

net electric field at origin, E = E₁ + E₂ + E₃ + E₄ ....

= k(2C)/1² - K(2C)/2² + k(2C)/3² - k(2C)/4² + ....

= 2K [ 1/1² - 1/2² + 1/3² - 1/4² + ..... ]

here $\Sigma^{\infty}_{k=1}\frac{(-1)^{(k+1)}}{k^2}=\frac{\pi^2}{12}$

= 2K [π²/12]

= π²k/6

= π² × 9 × 10^9/6

= 14.7894 × 10^9 N/C

≈ 14.8 × 10^9 N/C

Therefore the electric field intensity at origin is 14.8 × 10^9 N/C.