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.

1 answer

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.