Point charges of 2 μC and 3 μC are placed 1.2 m apart as shown.0.4 m 0.8 m2 μC 3 μCXWhat is the electric field strength at the point labelled X?A6 62 20 02 10 3 104π (0.4) 4π (0.8)ε ε− −× ×+B6 62 20 02 10 3 104π (0.4) 4π (0.8)ε ε− −× ×−C6 62 20 02 10 3 104π (0.8) 4π (0.4)ε ε− −× ×+D6 62 20 02 10 3 104π (0.8) 4π (0.4
Question
Point charges of 2 μC and 3 μC are placed 1.2 m apart as shown.0.4 m 0.8 m2 μC 3 μCXWhat is the electric field strength at the point labelled X?A6 62 20 02 10 3 104π (0.4) 4π (0.8)ε ε− −× ×+B6 62 20 02 10 3 104π (0.4) 4π (0.8)ε ε− −× ×−C6 62 20 02 10 3 104π (0.8) 4π (0.4)ε ε− −× ×+D6 62 20 02 10 3 104π (0.8) 4π (0.4
Solution
The electric field strength at a point due to a point charge is given by the formula:
E = kQ/r^2
where: E is the electric field strength, k is Coulomb's constant (9 x 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge.
The electric field strength at point X due to the 2 μC charge is:
E1 = kQ1/r1^2 = (9 x 10^9 N m^2/C^2) * (2 x 10^-6 C) / (0.4 m)^2
The electric field strength at point X due to the 3 μC charge is:
E2 = kQ2/r2^2 = (9 x 10^9 N m^2/C^2) * (3 x 10^-6 C) / (0.8 m)^2
The total electric field strength at point X is the vector sum of E1 and E2. Since the charges have the same sign, they will repel each other, and the electric field vectors will be in opposite directions. Therefore, the total electric field strength will be:
E_total = E1 - E2
This corresponds to option B in your question.
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