charges `2.0xx10^(-6)C` and `1.0 xx 10^(-6) C` are placed at. corners A and B of a square of side 5.0cm as shown in. figure .how much work will be done against. the electric field in moving a charge of `1.0 xx 10^(-6) C` from. C to D?.

Explanation

The electric potential at C
`= (1)/(4piepsilon_0) ((q_1 )/(AC)+ (q_2)/(BC))`
`= 9xx 10 ^9 Nm ^(-2)C(-2)((2.0xx10(-6)C)/(sqrt(2xx 0.05m) + (1.0xx10(-6)C)/(0.05m))`
`= (9000 V) ((2+sqrt2)/(sqrt2xx0.05))`
. The electric potential at D
`=(1)/(4piepsilon_0) ((q_1 )/(AD) + (q_2)/(BD)`
`= 9xx 10 ^9 Nm ^(-2)C(-2)((2.0xx10(-6)C)/( 0.05m) +(1.0xx10(-6)C)/(sqrt2xx0.05m))`
`= (9000 V) ((2sqrt2+1)/(sqrt2xx0.05))
. The work done against the electric field in moving the
charge `1.0xx 10(-6)C` frim C to d is `q(V_D- V_C)`
`= (1.0xx 10(-6) C)(900V) ((2sqrt2+1-2-sqrt2)/(sqrt2 xx o.05))`
=0.053J