Question

Identical point charges of +1.6 μC are fixed to diagonally opposite corners of a square. A third charge is then fixed at the center of the square, such that it causes the potentials at the empty corners to change signs without changing magnitudes. Find the sign and magnitude of the third charge.

Answer 1

To change the potentials at the empty corners of the square to opposite signs but equal magnitude, the third charge must be negative and calculated using the electric potential formula V = kQ/r, considering the superposition principle.

Step-by-step explanation:

To determine the sign and magnitude of the third charge that changes the potentials at the empty corners of the square without changing their magnitudes, we must consider the principle of superposition in electric potentials. Since identical point charges of +1.6 µC are fixed to diagonally opposite corners of a square, the potential due to these charges at the empty corners will be positive.

To change the sign without changing the magnitude, the third charge needs to have a negative sign and create an electric potential equal in magnitude but opposite in sign to the resultant potential created by the two fixed charges. The process involves calculating the potential due to the two charges at one of the empty corners and setting the potential due to the third charge, which is placed at the center of the square, to be equal in magnitude but negative in sign.

As the diagonally fixed charges are equidistant from any empty corner of the square, their potentials will add up. Therefore, the magnitude of the third charge can be calculated using the formula for the electric potential of a point charge, V = kQ/r, where k is Coulomb's constant, Q is the charge, and r is the distance from the charge to the point of interest. In this case, r would be the distance from the center of the square to any of its corners.

Answer 2

`q_(3) = - 4.8 * 10^(-6) C`

Step-by-step explanation:

given two charges

`q_(1) = q_(2)= q = +1.6 * 10^6 C \\`

let A , B, C, D are the corner where C and D are empty so we consider potential at C due to A and B is

using

`V = (kq)/(r)`

thus we have

`(V_(C))_(initial) = (kq)/(r)+(kq)/(r) \\(V_(C))_(initial) = 2(kq)/(r)` ...................(2)

let charge
`q_(3)` is placed at the center of two charge q as shown in figure

from Pythagoras we get

`d = √(( r^2+r^2))`

`(d)/(2) = (r)/(√(2))`

we get

`(V_(c))_(final) = 2(kq)/(r)+(kq_(3))/(d/2)`

`(V_(c))_(final) = 2(kq)/(r)+(k√(2) q_(3))/(r)` ........(3)

according to question such that it causes the potentials at the empty corners to change signs without changing magnitudes. from first and third equation

`(V_(C))_(initial) = -(V_(C))_(final)`

`we \ \ get \\(2kq)/(r) + (kq_(3)√(2))/(r) = - (2kq)/(r)`

on solving we get

`q_(3) = - 4.8 * 10^(-6) C`