A particle with a charge of 7.90 nC is in a uniform electric field directed to t

Question

A particle with a charge of 7.90 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved a distance of 9.00 cm , the additional force has done an amount of work equal to 7.40×105 J and the particle has kinetic energy equal to 4.35×105 J .

A.) What work was done by the electric force?

B.) What is the potential of the starting point with respect to the endpoint?

C.) What is the magnitude of the electric field?

Explanation / Answer

Wapplied = F(applied) x distance moved

W(electric) + W(applied) = delta KE

W(electric) + 7.4 x 10^-5 = 4.35 x 10^-5

W(electric) = - 3.05 x 10^-5J

Hence, W(electric) = -3.05 x 10^-5 J

b)we know that,

W(electric) = qV

V = W(electric)/q

V = 3.05 x 10^-5/7.9 x 10^-9 = 3.86 x 10^3 Volts

Hence, V = 3.86 x 10^3 Volts

C)E = V/d

E = 3.86 x 10^3/0.09 = 4.29 x 10^4 N/C

Hence, E = 4.29 x 10^4 N/C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.