Answer:
Q_2 = +/- 295.75*Q
Explanation:
Given:
- The charge of the first particle Q_1 = +Q
- The second charge = Q_2
- The position of first charge x_1 = 2a
- The position of the second charge x_2 = 13a
- The net Electric Field produced at origin is E_net = 2kQ / a^2
Find:
Explain how many values are possible for the unknown charge and find the possible values.
Solution:
- The Electric Field due to a charge is given by:
E = k*Q / r^2
Where, k: Coulomb's Constant
Q: The charge of particle
r: The distance from source
- The Electric Field due to charge 1:
E_1 = k*Q_1 / r^2
E_1 = k*Q / (2*a)^2
E_1 = k*Q / 4*a^2
- The Electric Field due to charge 2:
E_2 = k*Q_2 / r^2
E_2 = k*Q_2 / (13*a)^2
E_2 = +/- k*Q_2 / 169*a^2
- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:
E_net = E_1 + E_2
2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2
- The two equations are as follows:
1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2
2Q = Q / 4 + Q_2 / 169
Q_2 = 295.75*Q
2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2
2Q = Q / 4 - Q_2 / 169
Q_2 = -295.75*Q
- The two possible values corresponds to positive and negative charge Q_2.
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