Two point charges Q1 and Q2, are located a distance .20 meter apart, as shown above. Charge Q1 = +8.0μC. The net electric field is zero at point P, located .40 meter from Q1 and .20 meter from Q2.
Determine the magnitude and sign of charge Q2.
-0.02¦ÌC
To determine the magnitude and sign of charge Q2, we can make use of the fact that the net electric field at point P is zero.
The electric field due to Q1 at point P can be calculated using the formula:
E1 = k * Q1 / r1^2
where k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q1 is the charge of Q1, and r1 is the distance between Q1 and point P.
Similarly, the electric field due to Q2 at point P can be calculated using the same formula:
E2 = k * Q2 / r2^2
where Q2 is the charge of Q2, and r2 is the distance between Q2 and point P.
Since the net electric field at point P is zero, we can set up the following equation:
E1 + E2 = 0
Substituting the formulas for E1 and E2, we have:
k * Q1 / r1^2 + k * Q2 / r2^2 = 0
Now, we can plug in the given values:
Q1 = +8.0 μC (which is equivalent to 8.0 x 10^-6 C)
r1 = 0.40 m
r2 = 0.20 m
k = 9 x 10^9 Nm^2/C^2
Plugging in these values, we get:
(9 x 10^9 Nm^2/C^2) * (8.0 x 10^-6 C) / (0.40 m)^2 + (9 x 10^9 Nm^2/C^2) * Q2 / (0.20 m)^2 = 0
Simplifying the equation, we can isolate Q2:
Q2 = -0.25 Q1
Now, we substitute the given value of Q1:
Q2 = -0.25 * 8.0 x 10^-6 C
Calculating this expression, we find:
Q2 ≈ -2.0 x 10^-6 C
So, the magnitude of charge Q2 is approximately 2.0 μC, and its sign is negative.