Two point charges each of magnitude `2muC` are located at A and B. If AB=BC = CA = lmand AD = 0.5 m, find the amount of work done in transfering a test charge `q_0 = 2 muC` from D to C.

Explanation

In the given case, work done in moving a charge q from point D to C is (`V_( C) - V_(D))q_(0)`. VD is the potential at D due to the charges at points A and B. Vc is the potential at C due to the charges at points A and B. In the figure,

Now, potential at C due to the charge at A.
`V_(A) = q/(4pi epsilon_(0)r)`
Potential at C due to the charge at B.
`V_(B) = q/(4pi epsilon_(0)r)`
`therefore` Potential at C `=V_( C) = V_(A) + V_(B)`
`=q/(4pi epsilon_(0)r) + q/(4pi epsilon_(0)r) = (2q)/(4pi epsilon_(0)r)`
`=(2 xx (9 xx 10^(9)) xx (2 xx 10^(-6)))/1`
`=3.6 xx 10^(4) V`
Potential at D due to the charge at A,
`V_(A)^(.) q/(4pi epsilon_(0)(AD)) =q/(4pi epsilon_(0)(0.5))`
Potential at D due to the charge at B,
`V_(B)^(.) = q/(4pi epsilon_(0)(BD)) =q/(4pi epsilon_(0)(0.5))`
`therefore` Potential at D `=V_(D) = V_(A)^(.) + V_(B)^(.)`
`=q/(4pi epsilon_(0)(0.5)) + q/(4pi epsilon_(0)(0.5)) = (2q)/(4pi epsilon_(0)(0.5))`
`=7.2 xx 10^(4) V`
Work done, `W = q_(0)(V_( C) - V_( D))`
`=(3.6 xx 10^(4) - 72 xx 10^(4)) 2 xx 10^(-6)`
`=-7.2 xx 10^(-2) J = -72 mJ`