Question

Charges of \(-12.0 \mathrm{nC}\) and \(-22.0 \mathrm{nC}\) are separated by $0.700 \mathrm{m} .$ What is the potential midway between the two charges?

Short answer

Question: Determine the electric potential at the midpoint between two charges of \(-12.0 \, \mathrm{nC}\) and \(-22.0 \, \mathrm{nC}\) separated by a distance of \(0.700 \, \mathrm{m}\). Answer: The electric potential at the midpoint between the two charges is \(-876 \, \mathrm{V}\).

Step-by-step solution

Convert the charges to coulombs

We need to convert the charges from nanocoulombs to coulombs. Recall that 1 nC = \(10^{-9}\) C. \(-12.0 \, \mathrm{nC} = -12.0 \times 10^{-9} \, \mathrm{C}\) \(-22.0 \, \mathrm{nC} = -22.0 \times 10^{-9} \, \mathrm{C}\)

Find the distance to the midpoint

The distance between the charges is \(0.700 \, \mathrm{m}\), so the distance to the midpoint is half of that: \(r_{mid} = \frac{0.700}{2} \, \mathrm{m} = 0.350 \, \mathrm{m}\)

Find the electric potential due to each charge at the midpoint

We can use the formula \(V = k\frac{q}{r}\), where \(k\) is the electrostatic constant (\(k = 8.99 \times 10^9 \, \mathrm{Nm^2/C^2}\)). For the \(-12.0 \times 10^{-9} \, \mathrm{C}\) charge: \(V_1 = k\frac{q_1}{r_{mid}} = (8.99 \times 10^9 \, \mathrm{Nm^2/C^2})(-12.0 \times 10^{-9} \, \mathrm{C}) / (0.350 \, \mathrm{m})\) For the \(-22.0 \times 10^{-9} \, \mathrm{C}\) charge: \(V_2 = k\frac{q_2}{r_{mid}} = (8.99 \times 10^9 \, \mathrm{Nm^2/C^2})(-22.0 \times 10^{-9} \, \mathrm{C}) / (0.350 \, \mathrm{m})\)

Sum the electric potentials to find the total potential at the midpoint

We can simply add the potential due to each charge to find the total potential at the midpoint. \(V_{total} = V_1 + V_2\) Plug in the values we found in Step 3 and perform the calculation. \(V_{total} = (-3.08 \times 10^2 \, \mathrm{V}) + (-5.68 \times 10^2 \, \mathrm{V}) = -8.76 \times 10^2 \, \mathrm{V}\) So, the electric potential at the midpoint between the two charges is \(-876 \, \mathrm{V}\).