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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
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B
0.04V/m
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C
900V/m
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D
4500V/m
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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 $$D$$

2111604_125175_ans_5d0518a6fb5340f58ed865874c274373.png

Physics

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