Two positive charges Q1 and Q2 are placed 10 cm apart along the x-axis. Q1 is at
ID: 1391560 • Letter: T
Question
Two positive charges Q1 and Q2 are placed 10 cm apart along the x-axis. Q1 is at the origin with a value equal to 5 mu C and Q3 = 7 mu C. A test charge Qo = - 3 mu C is placed at 3.7 cm from Q1. The Electrostatic Potential Difference between point A and B is A small sphere of mass 5.1 g hangs vertically from an insulating thread. A horizontal electric field E is applied to the sphere and displaces it 10 degree from the vertical as shown. If E = 7.2 times 105 N/C the charge on the mass is [Hint: Use equilibrium of forces]Explanation / Answer
Given Q 1 = 5 x 10 -6 C
Q 2 = 7 x10 -6 C
Q o = -3 x 10 -6 C
Potential at point A is V = [KQ1 / 14 cm ] +[KQ2 / 4cm] +[KQo/10.3 cm ]
= [KQ1 / 0.14 m ] +[KQ2 / 0.04m] +[KQo/0.103 m ]
Where K = 8.99 x 10 9 Nm 2/ C 2
Substitute values you get V = (321071 ) + (1573250 ) - (261844)
= 1632476 volt
Similarly
Potential at point A is V ' = [KQ1 / 16 cm ] +[KQ2 / 6cm] +[KQo/12.3 cm ]
= [KQ1 / 0.16 m ] +[KQ2 / 0.06m] +[KQo/0.123 m ]
Where K = 8.99 x 10 9 Nm 2/ C 2
Substitute values you get V ' = 280937.5 + 1048833 - 219268
= 1110502.5 volt
Potential difference = V ' - V
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