Two positive charges Q1 and Q2 are placed 10 cm apart along the x-axis Q1 is at

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 Q2 = 7 mu C. A test charge Q0 = - 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 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

1)

Here , Potential due to a point charge is given as

V = kq/d

Now , at point A,

the potential , VA = 9 *10^9 *10^-6 * (5/0.14 - 3/0.103 + 7/.04)

VA = 1.634 *10^6 V

now , at point B,

VB = 9 *10^9 *10^-6 * (5/0.16 - 3/0.123 + 7/.06)

VB = 1.112 *10^6 V

the difference in the potential = (1.634 - 1.112) *10^6

the difference in the potential = 5.22 *10^5 V

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