Question

The electric potential at a distance of \(20.0 \mathrm{cm}\) from a point charge is \(+1.0 \mathrm{kV}\) (assuming \(V=0\) at infinity). (a) Is the point charge positive or negative? (b) At what distance is the potential $+2.0 \mathrm{kV} ?$

Short answer

Answer: The point charge is positive. The electric potential doubles to \(+2.0 \mathrm{kV}\) at a distance of \(10.0 \mathrm{cm}\) from the point charge.

Step-by-step solution

(a) Determine the sign of the point charge

We are given that the electric potential (\(V\)) at a distance of \(20.0 \mathrm{cm}\) is \(+1.0 \mathrm{kV}\). The electric potential will be positive when the charge (\(q\)) is positive and negative when the charge is negative. As \(V\) is positive, the point charge must be positive.

(b) Find the initial charge (\(q\))

To find the charge, we will use the formula \(V = \frac{k \cdot q}{r}\) and plug in the given values. Multiply both sides by \(r\) to isolate \(q\): \(q = \frac{V \cdot r}{k}\) Given \(V = 1.0 \times 10^3 \mathrm{V}\), \(r = 20.0 \mathrm{cm} = 0.2 \mathrm{m}\), and \(k = 8.99 \times 10^9 \frac{Nm^2}{C^2}\), plugging in these values, we get: \(q = \frac{(1.0 \times 10^3 \mathrm{V}) (0.2 \mathrm{m})}{8.99 \times 10^9 \frac{Nm^2}{C^2}}\) \(q = 2.22 \times 10^{-7} \mathrm{C}\) So, the charge is approximately \(2.22 \times 10^{-7}\) coulombs.

(c) Determine the distance at which the potential is \(+2.0 \mathrm{kV}\)

Now, we want to find the distance when the electric potential is doubled. Let's call it \(r_2\) and set the potential to \(2.0 \times 10^3 \mathrm{V}\): \(2.0 \times 10^3 \mathrm{V} = \frac{k \cdot q}{r_2}\) \(r_2 = \frac{k \cdot q}{2.0 \times 10^3 \mathrm{V}}\) Using the value of \(q\) found in step (b) and the known value of \(k\): \(r_2 = \frac{(8.99 \times 10^9 \frac{Nm^2}{C^2}) (2.22 \times 10^{-7}\mathrm{C})}{2.0 \times 10^3 \mathrm{V}}\) \(r_2 = 0.1 \mathrm{m}\) Thus, the electric potential is \(+2.0 \mathrm{kV}\) at a distance of \(0.1 \mathrm{m}\) or \(10.0 \mathrm{cm}\) from the point charge.