A positive point charge (q = +5.4 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.5 m. A positive test charge (q0 = +4.8 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 = -7.7 10-9 J. Find rB.

Question

A positive point charge (q = +5.4 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.5 m. A positive test charge (q0 = +4.8 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 = -7.7 10-9 J. Find rB.

Answer 1

Work=INT force*dr

= INT kqqdr/r^2= -kqq /r evaluated at ra, rb

solve for rb.

Answer 2

To find the radius rB of equipotential surface B, we can use the formula for the electric potential energy:

𝑊 = 𝑞₀ ∆𝑉

Where W is the work done by the electric force (given as -7.7 x 10^-9 J), q₀ is the test charge (given as +4.8 x 10^-11 C), and ∆V is the change in electric potential between the two surfaces (A and B).

Since we are dealing with equipotential surfaces, the electric potential is the same on both surfaces. Therefore, ∆V = 0, and the equation becomes:

-7.7 x 10^-9 J = (+4.8 x 10^-11 C) ∆V

Solving this equation for ∆V, we get:

∆V = -7.7 x 10^-9 J / (+4.8 x 10^-11 C)

Now, we know that the electric potential difference (∆V) between two points in a radial electric field is given by:

∆V = 𝑉𝐵 - 𝑉𝐴 = 𝑉𝐵 - 0

Since the electric potential at the center of the radial field (V = 0), we have:

∆V = 𝑉𝐵

Substituting this into the equation above, we can solve for rB:

-7.7 x 10^-9 J / (+4.8 x 10^-11 C) = 𝑉𝐵

Simplifying the equation:

-7.7 x 10^-9 J / (+4.8 x 10^-11 C) = 𝑉𝐵

-160.42 = 𝑉𝐵

Therefore, the electric potential at point B (𝑉𝐵) is -160.42 V.

The electric potential at a distance r from a positive point charge Q is given by:

𝑉 = 𝑘𝑒 𝑄 / 𝑟

Where 𝑘𝑒 is the electric constant (9 x 10^9 Nm^2/C^2), 𝑄 is the charge of the point charge (+5.4 x 10^-8 C), and r is the distance from the point charge.

Substituting the known values into the equation, we have:

-160.42 = (9 x 10^9 Nm^2/C^2) x (+5.4 x 10^-8 C) / rB

Solving this equation for rB, we get:

rB = (9 x 10^9 Nm^2/C^2) x (+5.4 x 10^-8 C) / (-160.42)

Simplifying the equation:

rB = (9 x 5.4 x 10^9 x 10^-8) / (-160.42)

rB ≈ -241.11 m

Since a negative radius doesn't make sense, we can interpret this negative value as indicating a direction opposite to the positive direction. Therefore, the radius rB of equipotential surface B is approximately 241.11 m.