In two opposite vertices of a square with sides a = 20 cm, there are opposite charges q1 = q2 = 50 nC.

| September 5, 2021 | education

In two opposite vertices of a square with sides a = 20 cm, there are opposite charges q1 = q2 = 50 nC. Determine the strength of the electrostatic field E in the center of the square.

To find the desired value of the tension in the center of the square under consideration, we apply the formula: E = E1 → + E2 → = 2E1 = 2 * k * q / (a * √2 / 2) ^ 2 = 4 * k * q / a2.

Constants and variables: k – coeff. proportionality (k = 9 * 10 ^ 9 m / F); q is the value of each of the opposite charges (q = 50 nC = 5 * 10 ^ -8 N); a – the length of the side of the square under consideration (a = 20 cm = 0.2 m).

Let’s make a calculation: E = 4 * k * q / a ^ 2 = 4 * 9 * 10 ^ 9 * 5 * 10 ^ -8 / 0.2 ^ 2 = 45 * 10 ^ 3 V / m = 45 kV / m.

Answer: The tension in the center of the square under consideration is 45 kV / m.



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