The centre of the ball, still carrying a charge of 1.2 × 10-6 C, is now placed 0.40 m froma point charge Q. The charge on the ball acts as a point charge at the centre of the ball.P is the point on the line joining the charges where the electric field strength is zero.The distance PQ is 0.22 m.0.40 m0.22 mP Q+1.2 µCCalculate the charge on Q. State your answer to an appropriate number ofsignificant figures.
Question
The centre of the ball, still carrying a charge of 1.2 × 10-6 C, is now placed 0.40 m froma point charge Q. The charge on the ball acts as a point charge at the centre of the ball.P is the point on the line joining the charges where the electric field strength is zero.The distance PQ is 0.22 m.0.40 m0.22 mP Q+1.2 µCCalculate the charge on Q. State your answer to an appropriate number ofsignificant figures.
Solution
The electric field at point P due to the two charges is zero. This means that the electric fields due to each charge at point P are equal in magnitude but opposite in direction.
The electric field E due to a point charge Q at a distance r is given by Coulomb's law:
E = kQ/r²
where k is Coulomb's constant (9.0 x 10^9 N m²/C²).
Let's denote the charge on the ball as Q1 (1.2 x 10^-6 C) and the charge we're trying to find as Q2. The distance from Q1 to P is 0.40 - 0.22 = 0.18 m, and the distance from Q2 to P is 0.22 m.
Setting the magnitudes of the two electric fields equal to each other gives us:
kQ1/(0.18 m)² = kQ2/(0.22 m)²
Solving for Q2 gives:
Q2 = Q1 * (0.22/0.18)²
Substituting Q1 = 1.2 x 10^-6 C gives:
Q2 = 1.2 x 10^-6 C * (0.22/0.18)² = 1.79 x 10^-6 C
So the charge on Q is approximately 1.79 µC.
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