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schepotkina [342]

3 years ago

14

A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106

N/C. If the test charge is replaced with another test charge of 23 mC, what happens to the external electric field at P

1 answer:

LenKa [72]3 years ago

8 0

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

$E_2 = \frac{E_1q_1}{q_2}$

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.

$E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C$

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.

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