A positive point charge (q = +5.62 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.50 m. A positive test charge (q0 = +4.91 x 10-11 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -8.54 x 10-9 J. Find rB.

Question

A positive point charge (q = +5.62 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.50 m. A positive test charge (q0 = +4.91 x 10-11 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -8.54 x 10-9 J. Find rB.

Answer 1

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Answer 2

To find the radius of equipotential surface B (rB), we can use the equation for the work done by the electric force.

The work done (W) by the electric force on a test charge moving between two points can be calculated using the equation:

W = q0 * (VB - VA)

where q0 is the test charge, VB is the potential at the point of surface B, and VA is the potential at the point of surface A.

Given that the work done (W) is -8.54 x 10^-9 J and the test charge (q0) is +4.91 x 10^-11 C, we can rearrange the equation to solve for the potential difference (VB - VA):

VB - VA = W / q0

Substituting the given values, we get:

VB - VA = (-8.54 x 10^-9 J) / (4.91 x 10^-11 C)
= -174.16 V

Now, since the point charge at surface A creates an equipotential surface, the potential at surface A (VA) is the same as the potential of the point charge. The potential of a point charge can be calculated using the equation:

V = k * (q / r)

where V is the potential, k is the Coulomb's constant (9 x 10^9 N*m^2/C^2), q is the charge, and r is the distance from the point charge.

Substituting the given values for surface A (rA = 2.50 m and q = +5.62 x 10^-8 C), we get:

VA = k * (q / rA)
= (9 x 10^9 N*m^2/C^2) * ((5.62 x 10^-8 C) / (2.50 m))
= 2.01 x 10^6 V

Now we can substitute the values of VA (-2.01 x 10^6 V) and VB - VA (-174.16 V) into the equation VB - VA = -174.16 V and solve for VB:

VB - 2.01 x 10^6 V = -174.16 V

VB = -174.16 V + 2.01 x 10^6 V
= 1.83584 x 10^6 V

Finally, we can calculate the radius of surface B (rB) using the equation for potential:

V = k * (q / r)

Rearranging the equation to solve for r, we get:

r = k * (q / V)

Substituting the values of q (5.62 x 10^-8 C) and V (1.83584 x 10^6 V), we get:

rB = (9 x 10^9 N*m^2/C^2) * ((5.62 x 10^-8 C) / (1.83584 x 10^6 V))
= 0.275 m

Therefore, the radius of equipotential surface B (rB) is 0.275 m.