Identical point charges of +2.1 ìC 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.
To find the sign and magnitude of the third charge, we can use the principle of superposition in electrostatics. This principle states that the electric potential at any point in space due to a system of charges is the algebraic sum of the potentials due to each individual charge.
Given that diagonally opposite corners of the square have identical point charges of +2.1 μC, the potential at any one of those corners can be calculated using Coulomb's law. We can then equate the potential at that corner to zero to satisfy the condition in the problem, i.e., the potential changes sign without changing magnitude.
Let's go step by step:
1. Calculate the potential due to one of the charges at each empty corner:
Using Coulomb's law, the electric potential V at a point due to a point charge q is given by:
V = k*q/r
where k is the Coulomb's constant (approximately 9 x 10^9 Nm^2/C^2) and r is the distance from the charge.
2. Consider one of the empty corners:
Let's calculate the potential at one of the empty corners of the square, caused by one of the charges on the diagonally opposite corners.
3. Consider the potential due to the first charge:
We know that the potential due to a charge at a corner is given by V = k*q/r. Substitute the values:
V1 = k*(2.1 μC)/(length of one side of the square)
4. Consider the potential due to the second charge:
The potential at the same empty corner due to the second charge will be the same, V2 = k*(2.1 μC)/(length of one side of the square).
5. Consider the potential at the empty corner caused by the third charge:
Since we know that the potential at this empty corner changes sign without changing magnitude, we can equate V1 and V2:
V1 = -V2
(k*(2.1 μC)/(length of one side of the square)) = - (k*(2.1 μC)/(length of one side of the square))
Now, solving this equation will give us the sign and magnitude of the third charge:
6. Solve the equation:
Cross-multiply and simplify the equation:
(k*(2.1 μC)/(length of one side of the square)) = - (k*(2.1 μC)/(length of one side of the square))
2.1 μC = -2.1 μC
From this equation, we can see that the solution is not physically possible. It means that there is no charge that can be placed at the center of the square to satisfy the condition stated in the problem.
Therefore, the answer is that there is no third charge that can be placed at the center of the square to cause the potentials at the empty corners to change signs without changing magnitudes.
q=2.1μC=2.1•10⁻⁶ C,
the side of square ia “a” .
The potential at the empty corner due to two charges
φ=φ₁+φ₂ = kq/a + kq/a= 2kq/a
The potential at the empty corner due to three charges
φ`=φ₁+φ₂ +φ₃= kq/a + kq/a + kq₀2/(a√2)= 2kq/a + kq₀2/(a√2).
By the data
φ`= - 2kq/a, =>
2kq/a + kq₀2/(a√2) = - 2kq/a
kq₀2/(a√2) = - 4kq/a
q₀=2√2q=2•1.41•2.1•10⁻⁶ =5.94•10⁻⁶ C