Four point charges qA = 2 μC, qB = –5 μC, qC = 2 μC, and qD = –5 μC arelocated at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Given: The four point charges q A = 2 μC , q B = − 5 μC , q C = 2 μC and q D = - 5μC are located at four corners of square ABCD of side 10 cm and the charge placed at the center is 1 μC .
Consider the figure below.
The force of attraction between two charges located at a distance r is given as,
F ( r ) = 1 4 π ε 0 | q 1 q 2 | r 2 .
By substituting the given values in the above equation, we get
F OA = 1 4 π ε 0 | q A × q OA 2 | = 1 4 π ε 0 | 2 × 1 OA 2 |
Similarly,
F OC = 1 4 π ε 0 | q c × q OC 2 | = 1 4 π ε 0 | 2 × 1 OC 2 |
From the figure, OA = OC , therefore,
F OA = − F OC F OA + F OC = 0
Similarly, the sum of forces due to other two charges is also zero.
Thus, the net force on the charge placed at center will be zero.
Given: The four point charges
Consider the figure below.
The force of attraction between two charges located at a distance
By substituting the given values in the above equation, we get
Similarly,
From the figure,
Similarly, the sum of forces due to other two charges is also zero.
Thus, the net force on the charge placed at center will be zero.
