Question

A source charge of 3 µC generates an electric field of 2.86 × 105 N/C at the location of a test charge.

What is the distance, to the nearest hundredth, of the test charge from the source charge?

k equals 8.99 times 10 Superscript 9 baseline N times Startfraction meters squared over Coulombs squared.

Answer 1

To find the distance \( r \) from the source charge to the test charge, we can use the formula for the electric field \( E \) generated by a point charge \( Q \):

\[
E = \frac{k |Q|}{r^2}
\]

where:
- \( E \) is the electric field (in N/C),
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( |Q| \) is the magnitude of the source charge (in C),
- \( r \) is the distance between the source charge and the test charge (in m).

We know:
- The source charge \( Q = 3 \, \mu\text{C} = 3 \times 10^{-6} \, \text{C} \)
- The electric field \( E = 2.86 \times 10^5 \, \text{N/C} \)

We can rearrange the formula to solve for the distance \( r \):

\[
r^2 = \frac{k |Q|}{E}
\]

To isolate \( r \), we take the square root:

\[
r = \sqrt{\frac{k |Q|}{E}}
\]

Now, plug in the values:

\[
r = \sqrt{\frac{(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2)(3 \times 10^{-6} \, \text{C})}{2.86 \times 10^5 \, \text{N/C}}}
\]

First, we calculate the numerator:

\[
(8.99 \times 10^9)(3 \times 10^{-6}) = 26.97 \times 10^3 = 2.697 \times 10^4 \, \text{N m}^2/\text{C}
\]

Now calculate \( r^2 \):

\[
r^2 = \frac{2.697 \times 10^4}{2.86 \times 10^5}
\]

Calculating the division:

\[
r^2 = \frac{2.697 \times 10^4}{2.86 \times 10^5} = \frac{2.697}{2.86} \times 10^{-1} \approx 0.09432 \, \text{m}^2
\]

Now take the square root to find \( r \):

\[
r \approx \sqrt{0.09432} \approx 0.307 \, \text{m}
\]

Rounding to the nearest hundredth:

\[
r \approx 0.31 \, \text{m}
\]

Thus, the distance of the test charge from the source charge is approximately \(\boxed{0.31}\) meters.