Two tiny objects with equal charges of 59.0 µC are placed at two corners of a square with sides of 0.355 m, as shown. How far above and to the left of the corner of the square labeled A would you place a third small object with the same charge so that the electric field is zero at A?
I would need to see where A is in relation to the two occupied corners .
We often have a problem helping with "as shown in the figure" questions, when there is no figure
1----------2
- -
- -
- -
- -
- -
3-----------4
Sorry this is a horrible square but I couldn't copy the image. The A is on corner 1 and the two charges are on corners 3 and 4.
The third charge must lie along the direction of the resultant force due to the charges at A and B. The force due to B has half the magnitude of the force due to A, and is directed 45 degree to the left of the vertical force due to the charge at A. I find the resultant to be inclined to the vertical by an angle of 14.64 degrees, which is arctan[(sqrt2/4)/(1 + sqrt2/4)]
For the distance of the third charge from corner A, match the force due to the third charge to the resultant force magnitude due to charges at A and B.
Could you please explain that more clearly?
To find the position of the third small object where the electric field is zero at point A, we can use the principle of superposition.
Step 1: Calculate the electric field due to each of the two identical charges at point A.
The electric field due to a point charge can be calculated using the formula:
Electric field (E) = (k * q) / r^2
where k is the electrostatic constant (8.99 x 10^9 N*m²/C²), q is the charge, and r is the distance between the point charge and the point where the electric field is being calculated.
In this case, we need to calculate the electric field at point A due to each of the two charges located at the corners of the square.
Electric field due to a charge at one corner:
E1 = (k * q) / r1^2
Electric field due to a charge at the opposite corner:
E2 = (k * q) / r2^2
Step 2: Add the electric fields vectorially to find the net electric field at point A.
Since the electric field is a vector quantity, we need to consider both the magnitude and direction to find the net electric field.
Net electric field at point A:
E_net = E1 + E2
To cancel out the net electric field at point A, the third small object must be placed such that the magnitude and direction of the electric field it produces cancels out the net electric field produced by the other two charges.
Step 3: Calculate the distance and direction to place the third object.
We can use the principle of superposition to determine the position. Since the two charges are located at the corners of a square, the horizontal and vertical distance from each corner to point A is equal.
Assuming the distance from each corner to point A is "d," the net electric field at point A due to the two charges is given by:
E_net = √[(E1)^2 + (E2)^2]
Now, let's calculate the values of E1 and E2 using the given values:
Electric field due to a charge at one corner:
E1 = (8.99 x 10^9 N*m²/C²) * (59.0 x 10^(-6) C) / (d^2)
Electric field due to a charge at the opposite corner:
E2 = (8.99 x 10^9 N*m²/C²) * (59.0 x 10^(-6) C) / (d^2)
Next, calculate the net electric field at point A by summing the magnitudes of E1 and E2:
E_net = √[(E1)^2 + (E2)^2]
Finally, adjust the position of the third object. Starting from the corner labeled A, move horizontally "x" units to the left and vertically "y" units above the corner of the square labeled A.
Therefore, the final answer is the position (-x, y).
Note: The exact numerical values of "d," "x," and "y" can only be determined by solving the equations obtained in the previous steps.