Two particles are fixed to an x axis: particle 1 of charge -7.93 x 10-7 C is at the origin and particle 2 of charge +7.93 x 10-7 C is at x = 14.6 cm. Midway between the particles, what is the magnitude of the net electric field?
To find the magnitude of the net electric field midway between the two particles, we can use the principle of superposition. The net electric field at this point is the vector sum of the electric fields generated by each particle.
The electric field generated by a point charge can be calculated using Coulomb's Law:
E = k * (|q| / r^2)
where E is the electric field, k is the Coulomb's constant (8.99 x 10^9 N·m^2/C^2), |q| is the absolute value of the charge, and r is the distance between the charge and the point where we want to calculate the field.
In this case, particle 1 has a charge of -7.93 x 10^(-7) C at the origin (x = 0) and particle 2 has a charge of +7.93 x 10^(-7) C at x = 14.6 cm.
First, let's calculate the electric field generated by particle 1 at the midpoint.
r1 = 0.5 * 14.6 cm = 7.3 cm = 0.073 m
E1 = k * (|q1| / r1^2) = 8.99 x 10^9 N·m^2/C^2 * (7.93 x 10^(-7) C / (0.073 m)^2)
Next, let's calculate the electric field generated by particle 2 at the midpoint.
r2 = 0.5 * 14.6 cm = 7.3 cm = 0.073 m
E2 = k * (|q2| / r2^2) = 8.99 x 10^9 N·m^2/C^2 * (7.93 x 10^(-7) C / (0.073 m)^2)
Finally, to find the net electric field, we need to calculate the vector sum of E1 and E2. Since the particles are fixed on the x-axis and have equal magnitudes of charge, the electric fields they generate have the same magnitude but opposite directions. Therefore, the net electric field at the midpoint is given by:
E_net = E1 - E2
Substituting the values we calculated for E1 and E2:
E_net = E1 - E2 = (8.99 x 10^9 N·m^2/C^2 * (7.93 x 10^(-7) C / (0.073 m)^2)) - (8.99 x 10^9 N·m^2/C^2 * (7.93 x 10^(-7) C / (0.073 m)^2))
Simplifying the equation will give us the magnitude of the net electric field at the midpoint between the particles.