Derive an expression for electric potential due to a point Charge. ABCD is a square measuring 0.1m on each side. Positive charges of 2nC, 4nC and 8nC are placed at point A, B and C respectively. Calculate the work done to transfer 1C of charge from point D to centre of square.

Explanation

Electric potential due to a point charge
According to the principle of superposition of potentials, the potential at D is

`V_(D)= (1)/(4pi epsi_(0)) [(q_(A))/(AD) + (q_(B))/(BD) + (q_(C ))/(CD)]`
Here, AD= CD= 0.1m
But `BD= sqrt2xx` side (`because BD` is diagonal of the square)
`=sqrt2 xx 0.1m`
`therefore V_(D)= 9 xx 10^(9) [(2 xx 10^(-9))/(0.1) + (4 xx 10^(-9))/(0.1 sqrt2) + (8 xx 10^(-9))/(0.1)]`
`=9 [20 + 20 sqrt2 + 80] = 1154.56V`
Potential at centre O of square ABCD,
`V_(0)= (1)/(4pi epsi_(0)) [(2 xx 10^(-9))/(0.1 sqrt2//2)+ (4 xx 10^(-9))/(0.1 sqrt2//2) + (8 xx 10^(-9))/(0.1 sqrt2//2)]`
`=9 [20 sqrt2 + 40sqrt2+ 80 sqrt2]`
`=9 xx 140 xx 2= 1781.91V`
Potential difference, `Delta V= V_(O)-V_(D)= 1781.91 - 1154.56= 627.35V`
`therefore` Word done= `Delta Vq= 627.35 xx 1= 627.35 J`