Consider four masses arranged (clockwise from A at the top) in a cross (looks like a giant plus symbol). Each is 1cm from the center of the cross. Mass A has a charge +1𝜇C , mass B has a charge of −1𝜇C , and mass C has a charge of +2𝜇C . At the center the electric field points 30 degrees (Northeast) from the vertical axis.
Is there a formula I need to find the missing value of D? I tried setting opposite ends equal to separate and find D but I cant figure this out or find anything online like it. Thank you.
look at the relative field strengths (at the center) from the individual charges
... they are all one cm away
... so the contribution to the net field is proportional to the charge
north-south is +1 N
east-west is ... D - 1 E
(E-W) / (N-S) = tan(30º)
I calculated a charge of -1.1547E-6
I used the (force between north and south and multiplied that by tan(30) ) and set that equal to (E*W) to solve for W. Is this what you were saying? Sorry I'm new to Physics so I'm trying to get this right to refer back to later on.
Thank you
I immediately realized how my interpretation & calculations were incorrect, but can you help elaborate further on how to solve for the value of the charge on D?
If you have time to write out the steps that would be greatly appreciated, I already lost my chance to earn credit for the problem but would love to know how to solve this.
Thank you.
i need help with this exact same question, could someone please answer?
To find the missing value D, we can use the principle of superposition in electrostatics. The electric field at the center of the cross is the vector sum of the electric fields created by each individual charge.
Let's consider the electric field created by each charge separately. The electric field created by a point charge is given by the formula:
E = k * (q / r^2)
where E is the electric field, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
Now, let's calculate the Electric field created by each charge at the center:
1. Charge A:
The electric field created by charge A at the center is given by:
E_A = k * (q_A / r_A^2)
Substituting the values: q_A = +1μC and r_A = 1cm = 0.01m, we get:
E_A = (8.99 x 10^9 Nm^2/C^2) * (1 x 10^-6 C) / (0.01m)^2
2. Charge B:
The electric field created by charge B at the center is given by:
E_B = k * (q_B / r_B^2)
Substituting the values: q_B = -1μC and r_B = 1cm = 0.01m, we get:
E_B = (8.99 x 10^9 Nm^2/C^2) * (-1 x 10^-6 C) / (0.01m)^2
3. Charge C:
The electric field created by charge C at the center is given by:
E_C = k * (q_C / r_C^2)
Substituting the values: q_C = +2μC and r_C = 1cm = 0.01m, we get:
E_C = (8.99 x 10^9 Nm^2/C^2) * (2 x 10^-6 C) / (0.01m)^2
Now, since the electric field at the center is the vector sum of the electric fields created by each charge, we can write:
E_total = E_A + E_B + E_C
Given that the electric field at the center points 30 degrees northeast from the vertical axis, we know the direction of the electric field but not the magnitude. Therefore, we can break down the total electric field into its vertical and horizontal components using trigonometry.
The vertical component of the electric field is given by:
E_vertical = E_total * sin(30 degrees)
The horizontal component of the electric field is given by:
E_horizontal = E_total * cos(30 degrees)
Since the vertical component is zero and the horizontal component is non-zero, we can equate them to find D, which represents the magnitude of the electric field at the center:
E_vertical = E_horizontal
0 = E_total * sin(30 degrees) - D
We know the value of E_total from our previous calculations, so we can solve for D.
D = E_total * sin(30 degrees)
By substituting the value of E_total, we can find the magnitude of the electric field at the center (D) using the above formula.