Question

Two tiny objects with equal charges of \(7.00 \mu \mathrm{C}\) are placed at the two lower corners of a square with sides of \(0.300 \mathrm{m},\) as shown. Find the electric field at point \(B\) midway between the upper left and right corners.

Short answer

Answer: The net electric field at point B is directed vertically downwards with a magnitude of \(4.68 \times 10^{4} N/C\).

Step-by-step solution

Determine the charges at point A and point C

The charges at both points A and C are equal, and their magnitude is given as \(7.00 \mu C\). Convert this into coulombs: \(7.00 \times 10^{-6} C\).

Compute the distance from the charges to point B

The distance from both A and C to point B can be calculated using the Pythagorean theorem. Because point B is midway between the upper left and right corners, we can find that the sides of the triangle formed between the lower charges and Point B (ACB) are equal to half of the square's side, \(0.300 m / 2 = 0.150 m\). Now, find the hypotenuse (length AB or CB) using the Pythagorean theorem: \(\sqrt{(0.150 m)^2 + (0.300 m)^2} \approx 0.335 m\).

Calculate the electric field due to each charge at point B

We can calculate the electric field due to each charge at point B using the formula for the electric field created by a point charge: \(\vec{E} = \frac{kQ}{r^2}\hat{r}\), where k is the electrostatic constant (\(k \approx 8.9875 \times 10^9 N m^2 C^{-2}\)), Q is the charge, and r is the distance between the charge and the point where we're finding the electric field. For both charges A and C, we have: \(\vec{E_A} = \frac{8.9875 \times 10^9 (7.00 \times 10^{-6})}{(0.335)^2}\hat{r_A} \approx 5.26 \times 10^{4} \hat{r_A} N/C \) \(\vec{E_C} = \frac{8.9875 \times 10^9 (7.00 \times 10^{-6})}{(0.335)^2}\hat{r_C} \approx 5.26 \times 10^{4} \hat{r_C} N/C\)

Determine the components of the electric field vectors

Now we need to express the electric field vectors in terms of their horizontal (\(x\)) and vertical (\(y\)) components. We can use trigonometry to do this, by calculating the angle between the electric field vector and the horizontal axis using \(tan(\theta) = \frac{0.150}{0.300}\). This gives us \(\theta \approx 26.565^\circ\). We can then find the components of each field as follows: \(\vec{E_Ax} = 5.26 \times 10^{4} cos(26.565^\circ) N/C \approx 4.69 \times 10^{4} N/C\) (to the right) \(\vec{E_Ay} = -5.26 \times 10^{4} sin(26.565^\circ) N/C \approx -2.34 \times 10^{4} N/C\) (downwards) \(\vec{E_Cx} = -5.26 \times 10^{4} cos(26.565^\circ) N/C \approx -4.69 \times 10^{4} N/C\) (to the left) \(\vec{E_Cy} = -5.26 \times 10^{4} sin(26.565^\circ) N/C \approx -2.34 \times 10^{4} N/C\) (downwards)

Determine the net electric field at point B

Now we can find the net electric field at point B by summing the components of the fields created by each charge: \(\vec{E_{net_x}} = \vec{E_{Ax}} + \vec{E_{Cx}} = 4.69 \times 10^{4} N/C - 4.69 \times 10^{4} N/C = 0 N/C\) \(\vec{E_{net_y}} = \vec{E_{Ay}} + \vec{E_{Cy}} = -2.34 \times 10^{4} N/C - 2.34 \times 10^{4} N/C \approx -4.68 \times 10^{4} N/C\) Thus, the net electric field at point B is \(\vec{E_{net}} = 0\hat{i} - 4.68 \times 10^{4}\hat{j} N/C\). It is directed vertically downwards with a magnitude of \(4.68 \times 10^{4} N/C\).