An electric field does 9.63 MeV of work in moving a very small charged particle

Question

An electric field does 9.63 MeV of work in moving a very small charged particle from point a to point b through a potential difference of -3.21 MV. Find the charge of the particle as a multiple of e. The electric potential is given by (25 V/m^7)(xz^6 + x^3y^4) in the region including the point (x, y, z) = (-4, 3, -2) m exactly. Find the z-component of the electric field at that point. SHOW ALL YOUR STEPS FOR CREDIT. As shown, one point charge of -4.00 nC is located at y = +5.00 mm and another point charge of +5.00 nC is located at y = -4.00 mm, . Find the electric potential energy of this system of two point charges. Calculate their electric potential at y = +1.00 mm on the line between the two point charges of Problem 3 above. Suppose we move along the x-axis from x_a = 0.0 m (where the potential is 770 V) to x_b = 1.5 m. Along the x-axis in this region, the electric field has a magnitude given by (88 V/m^4)x^3 and makes an angle of 36 87' with the +x-direction. Find the potential at x_b = 1.5 m. SHOW ALL YOUR STEPS FOR FULL CREDIT.

Explanation / Answer

1)   We know that    Work done = W = V*Q

9.63MeV = -Q*3.21 MV

Q = -(9.63/3.21) e = -3e

2)   V = 25*(xz^6 + x^3y^4)

     Now   Ez = -dV/dz = -150xz^5

     Ez = -150*(-4)*(-2)^5 = -19200 N/C

3) Electric potential energy = E = kQ1Q2/r = (9*10^9*4*10^-9*5*10^-9)/9*10^-3

E = 20*10^-6 J

4) V = 9*10^9[(-4*10^-9/4*10^-3) + (5*10^-9/5*10^-3)] = 0

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