Consider four masses arranged as shown. Each is 1 cm from the center of the cross. Mass A has a charge +1 uC, mass B has a charge of -1uC, and mass C has a charge of +2uC. At the center the electric field points 30 degrees from the vertical axis as shown.
To find the net electric field at the center, we need to find the electric field due to each individual charge and then add them up vectorially.
For a point charge, the electric field is given by the formula:
\[E = \frac{kQ}{r^2}\]
where E is the magnitude of the electric field, k is the Coulomb's constant (which is equal to 9 x 10^9 N m^2/C^2), Q is the charge of the point charge, and r is the distance from the point charge.
First, let's find the electric field due to mass A.
Given: Q_A = 1uC = 1 x 10^-6 C, r_A = 1 cm = 0.01 m.
Using the formula for electric field, we get:
\[E_A = \frac{(9 x 10^9)(1 x 10^-6)}{(0.01)^2} = (9 x 10^11) N/C\]
The electric field due to mass B will have the same magnitude but opposite direction since it has a negative charge.
So: E_B = -(9 x 10^11) N/C
Given: Q_C = 2uC = 2 x 10^-6 C, r_C = 1 cm = 0.01 m.
Using the formula for electric field, we get:
\[E_C = \frac{(9 x 10^9)(2 x 10^-6)}{(0.01)^2} = (1.8 x 10^12) N/C\]
The electric field due to mass D will also have the same magnitude but opposite direction since it has a negative charge.
So: E_D = -(1.8 x 10^12) N/C
Now, let's consider the direction of the net electric field at the center. Given that the electric field points 30 degrees from the vertical axis, we can draw an equilateral triangle with each side representing the magnitude of each electric field.
By symmetry, we can see that the net electric field will also point 30 degrees from the vertical axis, but in the opposite direction. This means the electric field due to masses A and B will cancel out each other along the vertical axis, leaving only the electric field due to masses C and D contributing to the net electric field at the center.
Adding up the electric fields due to masses C and D, we get:
\[E_{net} = E_C + E_D = (1.8 x 10^12) N/C + (-1.8 x 10^12) N/C = 0\]
Therefore, the net electric field at the center is zero.