A uniform electric field of 2.00 kN/C is in the x direction. Apositive point cha

Question

A uniform electric field of 2.00 kN/C is in the x direction. Apositive point charge Q=3.00 mC is released from rest at theorigin. A) What is the potential difference between the origin and4.00 m from the origin (V(4m)-V(0m))? B) What is the change in the potential energy of the chargefrom x=0m to x=4m for the charge. C) What is the kinetic energy of the charge if it moved fromx=0m to x=4m without any loss of energy? D) What is the potential V(x) as a function of x if V(x) iszero at x=0? A uniform electric field of 2.00 kN/C is in the x direction. Apositive point charge Q=3.00 mC is released from rest at theorigin. A) What is the potential difference between the origin and4.00 m from the origin (V(4m)-V(0m))? B) What is the change in the potential energy of the chargefrom x=0m to x=4m for the charge. C) What is the kinetic energy of the charge if it moved fromx=0m to x=4m without any loss of energy? D) What is the potential V(x) as a function of x if V(x) iszero at x=0?

Explanation / Answer

E = 2.00 kN/C. Q = 3.00 mC A) V(4m)-V(0m) = -E*(4 - 0) = -8.00 kV B) U = Q*[V(4m)-V(0m)] = -24 J C) K = -U = 24 J D) V(x) = -Ex = -2.00x kV

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