Question

A raindrop inside a thundercloud has charge - \(8 e\). What is the electric force on the raindrop if the electric field at its location (due to other charges in the cloud) has magnitude $2.0 \times 10^{6} \mathrm{N} / \mathrm{C}$ and is directed upward?

Short answer

Answer: The electric force on the raindrop is \(25.6 \times 10^{-13} \ N\) downward.

Step-by-step solution

Identify given quantities

We are given the charge q = -\(8 e\) and the electric field E = \(2.0 \times 10^6 \ N/C\).

Express the charge of the raindrop in Coulombs

The charge of an electron is \(e = 1.6 \times 10^{-19} \ C\). So, the raindrop has charge: $$ q = -8 \times 1.6 \times 10^{-19} \ C $$

Calculate the electric force

Now, we can use the relationship between electric force, charge, and electric field to find the force: $$ F = Eq =(-8 \times 1.6 \times 10^{-19} \ C) \times (2.0 \times 10^6 \ N/C) $$

Simplify to find the magnitude of the electric force

After multiplying the charge and electric field, we get: $$ F = - 8 \times 1.6 \times 10^{-19} \times 2.0 \times 10^{6} \ N = - 25.6 \times 10^{-13} \ N $$ Since the force is negative, it is directed downward.

State the final answer

The electric force on the raindrop is \(25.6 \times 10^{-13} \ N\) downward.