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.
What is the charge of mass D in uC?
To find the charge of mass D, we need to consider the electric field created by masses A, B, and C at the center.
Let's start with mass A (+1 uC). The electric field created by mass A at the center can be found using Coulomb's law:
\[E_A = \frac{k \cdot q_A}{r_A^2},\]
where \(k\) is the electrostatic constant (approximately \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\)), \(q_A\) is the charge of mass A (\(+1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C}\)), and \(r_A\) is the distance between mass A and the center point (1 cm = 0.01 m in this case).
Considering the symmetry of the system, the electric field created by mass A will have both horizontal and vertical components. The vertical component will balance out the vertical components of the electric fields created by masses B and C to create the overall 30 degree angle. The horizontal component will create the horizontal component of the electric field at the center.
Next, let's consider mass B (-1 uC). The electric field created by mass B at the center can be found using the same equation as above. However, for mass B, the sign of the charge will produce an electric field with an opposite direction:
\[E_B = \frac{k \cdot q_B}{r_B^2},\]
where \(q_B\) is the charge of mass B (\(-1 \, \mu\text{C} = -1 \times 10^{-6} \, \text{C}\)) and \(r_B\) is the distance between mass B and the center point (1 cm = 0.01 m in this case).
Similarly, let's consider mass C (+2 uC). The electric field created by mass C at the center can be found using the same formula:
\[E_C = \frac{k \cdot q_C}{r_C^2},\]
where \(q_C\) is the charge of mass C (\(+2 \, \mu\text{C} = 2 \times 10^{-6} \, \text{C}\)) and \(r_C\) is the distance between mass C and the center point (1 cm = 0.01 m in this case).
Since the electric fields created by masses A and B cancel out in the vertical direction, the total vertical electric field at the center is only contributed by mass C:
\[E_{\text{vertical}} = E_C = \frac{k \cdot q_C}{r_C^2}.\]
To find the horizontal component of the electric field at the center, we need to consider the forces created by masses A, B, and C on mass D. The net horizontal force on mass D must be zero for it to remain stationary. Since the electric field created by mass A has a component in the x-direction, it will create a horizontal force on mass D. Similarly, the horizontal component of the electric field created by mass C will create a force on mass D. In contrast, the horizontal components of the electric fields created by masses B and C will cancel each other out.
Therefore, the net horizontal force on mass D is given by:
\[f_{\text{horizontal}} = f_{A,\text{horizontal}} + f_{C,\text{horizontal}} = q_D \cdot E_{A,\text{horizontal}} + q_D \cdot E_{C,\text{horizontal}} = 0,\]
where \(f_{A,\text{horizontal}}\) and \(f_{C,\text{horizontal}}\) are the horizontal forces created by masses A and C on mass D, and \(E_{A,\text{horizontal}}\) and \(E_{C,\text{horizontal}}\) are the horizontal components of the electric fields created by masses A and C at the center, respectively. Since the magnitudes of \(E_{A,\text{horizontal}}\) and \(E_{C,\text{horizontal}}\) are the same, we can write:
\[E_{\text{horizontal}} = E_{A,\text{horizontal}} + E_{C,\text{horizontal}} = 0.\]
Since the total horizontal electric field at the center is 0, there is no net horizontal force on mass D. Therefore,
\[0 = q_D \cdot E_{\text{horizontal}} = q_D \cdot (E_{A,\text{horizontal}} + E_{C,\text{horizontal}}) = q_D \cdot (0) = 0.\]
This implies that mass D is an uncharged object, so the charge of mass D is 0 uC.