Related Problems with Solution
Problem 1 : Two point charges, Q1 and Q2, are placed 2 meters apart. Q1 is +4 μC and Q2 is -3 μC. Calculate the electric field at a point on the line joining the two charges and 1 meter from Q1.
Solution :
To find the electric field at the given point, we’ll calculate the electric field due to each charge individually and then combine them using vector addition.
Electric field due to Q1 at the given point: $$[E_1 = \frac{k \cdot |Q_1|}{r^2} = \frac{8.99 \times 10^9 \cdot 4 \times 10^{-6}}{(1)^2} = 35.96 , \text{N/C}] (directed to the right)$$
Electric field due to Q2 at the given point: $$[E_2 = \frac{k \cdot |Q_2|}{r^2} = \frac{8.99 \times 10^9 \cdot 3 \times 10^{-6}}{(1)^2} = 26.97 , \text{N/C}] (directed to the left)$$
Now, find the net electric field: $$[E_{\text{net}} = E_1 + E_2 = 35.96 , \text{N/C} - 26.97 , \text{N/C} = 8.99 , \text{N/C}]$$
So, the electric field at the given point is 8.99 N/C to the right.
