1. Calculate the electric field contribution from each positive charge separately using the formula:
E = k * (q / r^2)
where:
- E is the magnitude of the electric field
- k is the Coulomb's constant, approximately 9.0 x 10^9 Nm^2/C^2
- q is the charge of each positive charge, 2.0 mc
- r is the distance from the positive charge to the negative test charge, which is 1.4 m for this example
For each positive charge, the electric field at the position of the negative test charge is:
E1 = 9.0 x 10^9 Nm^2/C^2 * (2.0 x 10^-6 C) / (1.4 m)^2
= (2.57 x 10^6 N/C) / (1.96 m^2)
= 1.31 x 10^6 N/C
E2 = 9.0 x 10^9 Nm^2/C^2 * (2.0 x 10^-6 C) / (1.4 m)^2
= (2.57 x 10^6 N/C) / (1.96 m^2)
= 1.31 x 10^6 N/C
2. Calculate the net electric field vector at the position of the negative test charge by summing the contributions from the two positive charges:
Electric field at position of negative charge: E_net = E1 + E2
= 1.31 x 10^6 N/C + 1.31 x 10^6 N/C
= 2.62 x 10^6 N/C
3. The direction of the electric field vector at the position of the negative test charge can be determined based on the charges of the positive charges. Since the test charge is negative and the positive charges are positive, the electric field acts away from the positive charges and towards the negative test charge.
Therefore, the electric field vector at the position of the negative test charge is:
E_net = 2.62 x 10^6 N/C, pointing towards the negative test charge.
Acceleration of Point Charge Due to Two Other Point Charges
Example
A negative test charge sits 1.4 m away from the middle point between a pair of positive charges. The positive charges are separated by 1.0 m and each have a charge of 2.0 mc.
What is the electric field vector at the position of the negative test charge?
in simple steps
1 answer
1 answer