Two point charges q 1 = + 2.60 n C and q 2 = ? 6.20 n C are 0.100 m apart. Point

Question

Two point charges q1=+2.60nC and q2=?6.20nC are 0.100m apart. Point A is midway between them; point B is 8.00

Two point charges q1=+2.60nC and q2=?6.20nC are 0.100m apart. Point A is midway between them; point B is 8.00times10?2m from q1 and 6.00times10?2m from q2. (See the figure below(Figure 1) .) Take the electric potential to be zero at infinity. Find the potential at point A Find the potential at point B Find the work done by the electric field on a charge of 2.35nC that travels from point B to point A

Explanation / Answer

Electric potential due to point charge = (k *q) / r

k = 9 * 10^9

q = charge in Coulombs

r = distance for the charge in meters

Part A

Find the potential at point A

V=________V

Point A is 0.050 m from each charge. So, r = 0.05 m

For q1, V = (9 * 10^9 * 2.6 * 10-9 ) / 0.05 = 468

For q2, V = (9 * 10^9 * (-6.20 * 10-9)) / 0.05 = -1116

Total potential at point A = 468 + (-1116) = -648 volts

Part B

Find the potential at point B

V=________V

For q1, V = (9 * 10^9 * 2.6 * 10-9) / 0.08 = 292.5

For q2, V = (9 * 10^9 * -6.2 * 10-9) / 0.06 = -930

Total potential at point B = 292.5 + (-930)  = -637.5 volts

Part C

Find the work done by the electric field on a charge of 2.35nC that travels from point B to point A

W=________

Work = Absolute value of Change of Potential energy

Potential energy = Electric potential * charge

Change of Potential energy = Change of electric potential * charge

Change of electric potential = Potential at A

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