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What is the electric field strength at a distance of 200 mm from a charge of 2 x 10-6 Coulomb, in vacuum?
- a)450 kV/m
- b)236 kV/m
- c)525 kV/m
- d)328 kV/m
Correct answer is option 'A'. Can you explain this answer?
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The electric field strength at a distance from a charge can be calculated using the formula:
\[
E = \frac{Q}{{4\pi\epsilon_0 r^2}}
\]
Where:
- E is the electric field strength
- Q is the charge
- \(\epsilon_0\) is the permittivity of free space
- r is the distance from the charge
Given:
- Q = 2 x 10^(-6) Coulomb
- r = 200 mm = 0.2 m (since 1 mm = 0.001 m)
We can substitute these values into the formula to calculate the electric field strength.
\[
E = \frac{{2 \times 10^(-6)}}{{4\pi\epsilon_0 \times (0.2)^2}}
\]
The permittivity of free space, \(\epsilon_0\), is a constant equal to 8.854 x 10^(-12) F/m.
\[
E = \frac{{2 \times 10^(-6)}}{{4\pi \times 8.854 \times 10^(-12) \times (0.2)^2}}
\]
Simplifying the expression further:
\[
E = \frac{{2 \times 10^(-6)}}{{4 \times 3.14159 \times 8.854 \times 10^(-12) \times 0.04}}
\]
\[
E = \frac{{2 \times 10^(-6)}}{{4 \times 3.14159 \times 8.854 \times 10^(-14)}}
\]
\[
E = \frac{{2 \times 10^(-6)}}{{1.1131 \times 10^(-12)}}
\]
\[
E = \frac{{2}}{{1.1131}} \times 10^(-6+12)
\]
\[
E = 1.795 \times 10^6 V/m
\]
Converting this value to kilovolts per meter (kV/m):
\[
E = 1.795 \times 10^3 kV/m
\]
Rounding off to the nearest whole number, the electric field strength at a distance of 200 mm from the charge is approximately 1.8 kV/m.
Therefore, none of the given options match the correct answer.
\[
E = \frac{Q}{{4\pi\epsilon_0 r^2}}
\]
Where:
- E is the electric field strength
- Q is the charge
- \(\epsilon_0\) is the permittivity of free space
- r is the distance from the charge
Given:
- Q = 2 x 10^(-6) Coulomb
- r = 200 mm = 0.2 m (since 1 mm = 0.001 m)
We can substitute these values into the formula to calculate the electric field strength.
\[
E = \frac{{2 \times 10^(-6)}}{{4\pi\epsilon_0 \times (0.2)^2}}
\]
The permittivity of free space, \(\epsilon_0\), is a constant equal to 8.854 x 10^(-12) F/m.
\[
E = \frac{{2 \times 10^(-6)}}{{4\pi \times 8.854 \times 10^(-12) \times (0.2)^2}}
\]
Simplifying the expression further:
\[
E = \frac{{2 \times 10^(-6)}}{{4 \times 3.14159 \times 8.854 \times 10^(-12) \times 0.04}}
\]
\[
E = \frac{{2 \times 10^(-6)}}{{4 \times 3.14159 \times 8.854 \times 10^(-14)}}
\]
\[
E = \frac{{2 \times 10^(-6)}}{{1.1131 \times 10^(-12)}}
\]
\[
E = \frac{{2}}{{1.1131}} \times 10^(-6+12)
\]
\[
E = 1.795 \times 10^6 V/m
\]
Converting this value to kilovolts per meter (kV/m):
\[
E = 1.795 \times 10^3 kV/m
\]
Rounding off to the nearest whole number, the electric field strength at a distance of 200 mm from the charge is approximately 1.8 kV/m.
Therefore, none of the given options match the correct answer.
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