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A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2 xx 10^(-9)C from a point P(0,0,3 cm) to a point Q10,4 cm, 0) via a point R(0,6 cm, 9 cm). |
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Answer» <P> Solution :As electrostatic field is a CONSERVATIVE field HENCE work done in taking a small charge from point P to Q is independent of the path followed.As q = 8 mC `= 8 XX 10^(-3) C, q_0 = -2 xx 10^(-9) C` At point P, `r_1 = 3 cm= 0.03 m` and at point Q,`r_2= 4 cm = 0.04 m` `:.` Work done `W = q_0 (V_2 - V_1) = q_0 1/(4pi epsi_0) .q[1/r_2-1/r_1] = (q_0q)/(4pi epsi_0) [ 1/r_2 - 1/r_1]` `:. W = 9 xx 10^9 xx (-2 xx 10^(-9))xx (8 xx 10^(-3)) [ 1/0.04 - 1/0.03] = 1.2 J` |
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