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Two point charges +15 μC and -10 C are separated by a distance of 20 cm in air. Calculate the electric field at the mid point of line joining two charges. If a point charge of 20 mC is placed at that mid-point, What is the magnitude of electric force experienced by it?
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Two point charges +15 μC and -10 C are separated by a distance of 20 c...
Electric Field Calculation at Mid-Point:
- Given charges: +15 μC and -10 μC
- Distance between charges: 20 cm = 0.2 m
- Electric field at mid-point can be calculated using the formula:
E = (k * |q1| / r1^2) + (k * |q2| / r2^2)
where k is the Coulomb's constant (8.9875 x 10^9 N m^2/C^2)
- Substituting the values, we get:
E = (8.9875 x 10^9 * 15 x 10^-6) / (0.1)^2 + (8.9875 x 10^9 * 10 x 10^-6) / (0.1)^2
E = 674.06 N/C + 449.37 N/C
E = 1123.43 N/C
Electric Force Calculation on 20 mC charge:
- Given charge at mid-point: 20 mC = 20 x 10^-3 C
- Electric field at mid-point: 1123.43 N/C
- Electric force experienced by the charge can be calculated using the formula:
F = q * E
where q is the charge and E is the electric field
- Substituting the values, we get:
F = 20 x 10^-3 * 1123.43
F = 22.47 N
Therefore, the magnitude of the electric force experienced by the 20 mC charge placed at the mid-point of the line joining the two charges is 22.47 N. This force is attractive if the charge is negative and repulsive if the charge is positive, as it interacts with the electric field created by the two point charges.
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Two point charges +15 μC and -10 C are separated by a distance of 20 c...
22.7
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Two point charges +15 μC and -10 C are separated by a distance of 20 cm in air. Calculate the electric field at the mid point of line joining two charges. If a point charge of 20 mC is placed at that mid-point, What is the magnitude of electric force experienced by it?
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