Question

Charge Q acts as a point charge to create an electric field it's strength measured at a distance of 30 cm away is 40N/C the magnitude of electric field strength that you would expect to measured at a distance of 60 cm away would be ?

• A. 10 N/C
• B. 160 N/C
• C. 4.4 N/C
• D. 400 N/C

Answer 1

Answer: (A)

10 N/C

Answer 2

The electric field strength ($E$) due to a point charge ($Q$) at a distance ($r$) is given by the formula:

$E = k \frac{Q}{r^2}$

where $k$ is Coulomb's constant.

This formula shows that the electric field strength is inversely proportional to the square of the distance from the point charge ($E \propto \frac{1}{r^2}$).

Given:

• Initial distance, $r_1 = 30 \text{ cm}$
• Initial electric field strength, $E_1 = 40 \text{ N/C}$
• Final distance, $r_2 = 60 \text{ cm}$

We want to find the electric field strength $E_2$ at distance $r_2$.

Using the proportionality, we can set up a ratio:

$\frac{E_2}{E_1} = \frac{k \frac{Q}{r_2^2}}{k \frac{Q}{r_1^2}}$

$\frac{E_2}{E_1} = \frac{r_1^2}{r_2^2}$

$\frac{E_2}{E_1} = \left(\frac{r_1}{r_2}\right)^2$

Substitute the given values:

$\frac{E_2}{40 \text{ N/C}} = \left(\frac{30 \text{ cm}}{60 \text{ cm}}\right)^2$

$\frac{E_2}{40 \text{ N/C}} = \left(\frac{1}{2}\right)^2$

$\frac{E_2}{40 \text{ N/C}} = \frac{1}{4}$

Now, solve for $E_2$:

$E_2 = 40 \text{ N/C} \times \frac{1}{4}$

$E_2 = 10 \text{ N/C}$

The magnitude of the electric field strength at a distance of 60 cm away would be 10 N/C.