Two point charges A = +3 text{ nC} and B = +1 | vedclass

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(A) Given: $q_A = 3 	imes 10^{-9} 	ext{ C}$,$q_B = 1 	imes 10^{-9} 	ext{ C}$,initial distance $d_i = 5 	imes 10^{-2} 	ext{ m}$.Initial potential

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$1.35 	imes 10^{-7} 	ext{ J}$

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(A) Given: $q_A = 3 	imes 10^{-9} 	ext{ C}$,$q_B = 1 	imes 10^{-9} 	ext{ C}$,initial distance $d_i = 5 	imes 10^{-2} 	ext{ m}$.Initial potential energy $U_i = \frac{1}{4\pi\epsilon_0} \frac{q_A q_B}{d_i} = (9 	imes 10^9) \frac{(3 	imes 10^{-9})(1 	imes 10^{-9})}{5 	imes 10^{-2}} = \frac{27 	imes 10^{-9}}{5 	imes 10^{-2}} = 5.4 	imes 10^{-7} 	ext{ J}$.After moving charge $B$ towards $A$ by $1 	ext{ cm}$,the new distance $d_f = 4 	imes 10^{-2} 	ext{ m}$.Final potential energy $U_f = \frac{1}{4\pi\epsilon_0} \frac{q_A q_B}{d_f} = (9 	imes 10^9) \frac{(3 	imes 10^{-9})(1 	imes 10^{-9})}{4 	imes 10^{-2}} = \frac{27 	imes 10^{-9}}{4 	imes 10^{-2}} = 6.75 	imes 10^{-7} 	ext{ J}$.Work done $W = U_f - U_i = (6.75 - 5.4) 	imes 10^{-7} 	ext{ J} = 1.35 	imes 10^{-7} 	ext{ J}$.

Two point charges $A = +3 \text{ nC}$ and $B = +1 \text{ nC}$ are placed $5 \text{ cm}$ apart in air. The work done to move charge $B$ towards $A$ by $1 \text{ cm}$ is

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