A metal sphere of diameter 5 cm holds a charge of -6 μC - Leaving Cert Physics - Question 9 - 2022

To find the electric field strength (E) at a distance of 3 cm from the surface of the sphere, we first need to calculate the distance from the center of the sphere:

• The radius of the sphere is half of the diameter:

  $r_{sphere} = \frac{5 \text{ cm}}{2} = 2.5 \text{ cm}$

• Therefore, at a distance of 3 cm from the surface, the total distance (d) from the center is:

  $d = r_{sphere} + 3 \text{ cm} = 2.5 \text{ cm} + 3 \text{ cm} = 5.5 \text{ cm} = 0.055 \text{ m}$

Using the formula for electric field strength:

$E = \frac{q}{4\pi \epsilon_0 r^2}$

where:

• $q = -6 \times 10^{-6} \text{ C}$ (the charge),
• $\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2$,
• $r = 0.055 \text{ m}$ (the total distance from the center).

Substituting the values into the equation gives:

$E = \frac{6 \times 10^{-6}}{4\pi (8.854 \times 10^{-12})(0.055^2)}$

Calculating this value yields:

$E \approx 1.78 \times 10^7 \text{ N/C}$

Thus, the electric field strength at a distance of 3 cm from the surface of the sphere is approximately $1.78 \times 10^7 \text{ N/C}$.