Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁
[tex]E_2 = \frac{E_1q_1}{q_2}[/tex]
where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.
[tex]E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C[/tex]
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.
