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ive tried a lot of methods but im wrong. even my tutor couldnt help me?
Identical point charges of +6.1 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.
1 Answer
- ?Lv 76 years ago
The potential at an empty corner of the square (of side a) before insertion of the third (center) charge is
V = k q / a + k q / a = 2kq/a
where k = 1/(4 pi eps0) and q is the charge at one of the corners.
The insertion of a charge Q at the center adds a contribution
k Q / (a/sqrt(2))
This additional contribution should be -2 * the original potential before the introduction of Q in order to flipthe sign of the potential at an empty corner:
k Q /(a/sqrt(2)) = -2 * (2 k q / a)
k Q / a = -4/sqrt(2) * k q /a
Comparing gives
Q = -4/sqrt(2) * q
= -2 sqrt(2) q
= - 2.83 * 6.1μC
= -17 μC