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Charges 2 xx 10^(-6)"C" and 1 xx 10^(-6)...

Charges `2 xx 10^(-6)"C" and 1 xx 10^(-6)`C are placed at comers A and B of a square of side 5 cm as shown in the figure. How much work will be done against the electric field in moving a charge of `1 xx 10^(-6)"C"` from C to D?

Text Solution

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The electric potential at C
`V_( C ) = (1)/( 4pi epsilon_(0) ) ((q_1 )/( AC ) + (q_2)/( BC) )`
From the diagram
`AC = sqrt( AB^(2) + BC^(2) ) = sqrt( 5^(2) + 5^(2) ) = sqrt(50)`
`AC = 5 sqrt2" cm" = 0.05 sqrt(2) m`
`BC = 5" cm" = 0.05" m"`
`therefore V_(C ) = 9 xx 10^(9) [ (2 xx 10^(-6) )/( 0.05 sqrt2) + ( 1xx 10^(-6) )/( 0.05) ]`
`= (9 xx 10^9 xx 10^(-6) )/( 0.05 ) [ (2)/( sqrt2) + 1]`
`V_( C ) = 434.6 xx 10^(3) V`
The electric potential at D
`V_( D) = (1)/( 4pi epsilon_(0) ) [ (q_1)/( AD) + (q_2)/( BD) ]`
`= 9 xx 10^(9) [ (2 xx 10^(-6) )/( 0.05) + (1 xx 10^(-6) ) /( 0.05 sqrt2) ]`
`= (9 xx 10^(9) xx 10^(-6) )/( 0.05 ) [ 2 + (1)/( sqrt2) ]`
`V_( D) = 487.3 xx 10^(3) ` V
The work done against the electric field in moving the charge of `1 xx 10^(-6)` C from C to D is
`W = q (V_( D) V_( C) )`
`W = 1 xx 10^(-6) [ 487.3 - 434.6] xx 10^(3)`
`W = 52.7 xx 10^(-3)`
`W = 0.0527` J
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