Question

A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitude of the electric force between these two charges is 7980000 N. What is the average separation between these charges

a/ 1.51 km
b/ 2.53 Km
c/ 1.51 m
d/ 1001 m

Answer 1

Answer: 1.51 km

Step-by-step explanation:

Coulomb's Law: The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,
`\vec{F}=k (Q_(1) Q_(2))/(r^(2))`

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

Given:

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

So,

`\begin{aligned}&7.98 * 10^(6)=\left(8.99 * 10^(9) \mathrm{Nm}^(2) / \mathrm{C}^(2)\right) \frac{48.8 \mathrm{C} * 41.7 \mathrm{C}}{\mathrm{r}^(2)} \\&r=\sqrt{(1.8294 * 10^(13))/(7.98 * 10^(6))}=1.514 * 10^(3) \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}`

Therefore, the separation between the two charges (r) = 1.51 km