Two point charges are fixed in place on an x axis. Particle 1 of charge q1 = q i

Question

Two point charges are fixed in place on an x axis. Particle 1 of charge q1 = q is located at x = 0, while particle 2 of charge q2 = -4q is located at x = +d. Consider a point P located on the x axis at x = +3d. Let q = 10 microcoulomb and d = 0.5 m. What is the electric potential at point P?

Two point charges are fixed in place on an x axis. Particle 1 of charge q1 = q is located at x = 0, while particle 2 of charge q2 = -4q is located at x = +d. Consider a point P located on the x axis at x = +3d. A particle of charge +q is moving along the x axis in the negative x direction. If its kinetic energy at point P is 1 J, what is its kinetic energy at x = 2d. Let q = 10 microcoulomb and d = 0.5 m.

Explanation / Answer

q1 = 10 uC
q2 = - 4*10 uC = - 40 uC

Kinetic Energy @ Point p = 1 J
Potential Energy @ Point p = k*q1*q/r1 - k*q2*q/r2
Potential Energy @ Point p = (8.9*10^9)* [(10*10)/(3*0.5) - (40*10)/(2*0.5)] * 10^-6 * 10^-6
Potential Energy @ Point p = - 2.97 J

At x = 2d
Potential Energy @ x = 2d = k*q1*q/r - k*q2*q/r
Potential Energy @ x = 2d = (8.9*10^9)* [(10*10)/(2*0.5) - (40*10)/(0.5)] * 10^-6 * 10^-6
Potential Energy @ Point p = - 6.23 J

Kinetic Energy = ?

Using Energy Conservation,
K.Ein + P.Ein = K.Efi + P.Efi
1 - 2.97 = K.Efin - 6.23
K.E fin = 4.26 J
Kinetic Energy @ x = 2d, K.E = 4.26 J

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