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. 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.

Explanation / Answer

Q1.

as work done=potential difference*charge

==>mangitude of charge=9.63 MeV/(3.21 MV)

=3e

Q2. electric field along z direction=-dV/dz

=25*(6*x*z^5)

so at x=-4 and z=-2

z component of electric field

=150*(-4)*(-2)^5=19200 V/m

Q3. electric potential energy of the system

=k*q1*q2/d

where d is the distance between two charges q1 and q2

k=coloumb's constant=9*10^9

so here potential energy=9*10^9*(-4*10^(-9))*5*10^(-9)/(9*0.001)

=-2*10^(-5) J

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