A small object carrying a charge of -5.50 nC is acted upon by a downward force o

Question

A small object carrying a charge of -5.50 nC is acted upon by a downward force of 30.0 nN when placed at a certain point in an electric field.

What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?

Three point charges with values q1 = 4.00 nC , q2 = 1.00 nC , and q3 = 7.00 nC are placed on three consecutive corners of a square whose side measures s = 5.00 m . Point A lies at the fourth corner of the square, diagonally across from q2. Point B lies at the center of the square. What are the values of the electric potential at point A and at point B? How much energy (supplied as work) is required to move a test charge q4 = 2.00 nC from point A to point B? Assume the zero of electric potential is chosen to be at infinity.

What is the electric potential at point A?

Explanation / Answer

F1 = E*q1

F2 = E*q2

F2/F1 = q2/q1

F2/(30*10^-9) = (1.6*10^-19)/(5.5*10^-9)

F = 8.73*10^-19 N

direction upwards

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VA = k*q/l + k*q3/l + k*q2/l*sqrt2

vA = (9*10^9)*((4/5) + (7/5) + (1/(5*sqrt2)))*10^-9

vA = 21.1 V

r = l/sqrt2 = 5/sqrt2

VB = k*q/r + k*q3/r + k*q2/r

vB = (9*10^9*sqrt2)*((4/5) + (7/5) + (1/5))*10^-9

vB = 30.55 V

W = (vB - vA)*q4 = 1.88*10^-8 J

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