Point charges, Q1=+44 nC and Q2=-72 nC, are placed as shown in Figure. An external force transports an electron from point A to point B. The work (in eV) done by the external force is...
Diagram is here: http://imgur.com/upJLx
Diagram is here: http://imgur.com/upJLx
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Hello Robert Moran, the basic technique is to find the difference in potential at points B and A
The basic formula for potential due to charge Q at distance r is K Q/r and K = 9*10^9
Potential at A due to the two charges
V1 = K 44*10^-9 / 0.6 volt and -K 72*10^-9 / 1 {because AQ2 = ./ (0.6)^2 + (0.8)^2}
Net potential due to Q1 and Q2 at A = 9 (220/3 - 72) V = 12 V
Same way the potential at B due to Q1 and Q2 will be got as 9 (44 - 120) V =
Now the potential difference between B and A, dV is (-684-12) = -696V
Hence the difference in potential energy = q dV
So the work done by the external force has to be 696 eV since q = -e
Negative sign indicates that the work is performed by the system.
The basic formula for potential due to charge Q at distance r is K Q/r and K = 9*10^9
Potential at A due to the two charges
V1 = K 44*10^-9 / 0.6 volt and -K 72*10^-9 / 1 {because AQ2 = ./ (0.6)^2 + (0.8)^2}
Net potential due to Q1 and Q2 at A = 9 (220/3 - 72) V = 12 V
Same way the potential at B due to Q1 and Q2 will be got as 9 (44 - 120) V =
Now the potential difference between B and A, dV is (-684-12) = -696V
Hence the difference in potential energy = q dV
So the work done by the external force has to be 696 eV since q = -e
Negative sign indicates that the work is performed by the system.