Question

# A point charge $$50\mu C$$ is located in the $$XY$$ plane at the point of position vector $${ \vec { r } }_{ 0 }=2\hat { i } +3\hat { j }$$. What is the electric field at the point vector $${ \vec { r } }_{ 1 }=8\hat { i } -5\hat { j }$$

A
1200V/m
B
0.04V/m
C
900V/m
D
4500V/m

Solution

## The correct option is D $$4500V/m$$$$\textbf{Step 1: Distance of charge from the point at which we have to find electric field [Refer Fig. 1]}$$Position vector of Charge Q,  $$\vec{r_0} = 2 \hat{i} + 3 \hat{j} ,$$              Position vector of Point where Electric Field has to be calculated,  $$\vec{r_1} = 8 \hat{i} - 5 \hat{j}$$ So, Distance between Point and Charge       $$\vec r = |\bigtriangleup\vec{r}| = |\vec{r_1} - \vec{r_0}| = |8 \hat{i} - 5 \hat{j} - 2 \hat{i} - 3 \hat{j}|$$                                                  $$= |6 \hat{i} - 8 \hat{j}|$$                     $$r = \sqrt{6^2 + 8^2} = 10m$$ $$\textbf{Step 2: Magnitude of Electric field [Refer Fig. 2]}$$                  $$Q = 50\ \mu C$$By Coulomb's Law, Electric field of a point charge at a distance $$r$$                   $$E = \dfrac{kQ}{r^2}$$                        $$= \dfrac{9 \times 10^{9} \times (50 \times 10^{-6}C)}{(10m)^2} = 4500\ V/m$$Hence Correct Option id $$D$$Physics

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