Question

Calculate the charge if the force is 5×10−6N, potential difference is 300V, and length is 15cm.

a) 1.5C
b) 2.0C
c) 3.0C
d) 5.0C

Answer 1

The formula relating force (F), potential difference (V), and length (L) in an electric field is given by:

\[ Q = F \times V \times L \]

Where:

- \( Q \) is the charge in coulombs (C),

- \( F \) is the force in newtons (N),

- \( V \) is the potential difference in volts (V), and

- \( L \) is the length in meters (m).

Given that the force is \( 5 \times 10^{-6} \, \text{N} \), potential difference is \( 300 \, \text{V} \), and length is \( 0.15 \, \text{m} \) (converted from 15 cm to meters), you can substitute these values into the formula:

\[ Q = (5 \times 10^{-6} \, \text{N}) \times (300 \, \text{V}) \times (0.15 \, \text{m}) \]

Now, calculate the charge:

\[ Q = 7.5 \times 10^{-4} \, \text{C} \]

So, the answer is:

c) 3.0C

Answer 2

To calculate the charge, multiply the potential difference by the force and the length.

Step-by-step explanation:

To calculate the charge, we can use the formula Q = V * F * d, where Q is the charge, V is the potential difference, F is the force, and d is the length. Plugging in the values given, we get Q = 300V * 5x10^-6N * 0.15m = 0.0225C. Therefore, the charge is 0.0225C, which is not one of the options given.