A point charge of 50 μC is placed in the XY plane at a location with radius vector r→_0=2i^+3j^ m The electric field strength and its magnitude at a point with radius vector r→=8i^-5j^ m is ε_o=8.85×10^(-12) C^2 N-m^(-2)

Correct answer: (A) 4.5 kV m^(-1). Separation vector, =r→-r_0→ =8i^-5j^-2i^+3j^ =6i^-8j^ r→-r→_0=√(6^2+-8^2)=10 Electric field strength, E=1/(4πε_0)qr^2 =1/(4×π×8.85×10^(-12))×(50×10^(-6))/100 =4495.9 =4.495×10^3≈4.5 kV m^(-1)

Q145 / 160 · TS EAMCET 2018 (04 May Shift 1) · Physics

A point charge of $50 μ C$ is placed in the $X Y$ plane at a location with radius vector ${\vec{r}}_{0} = 2 \hat{i} + 3 \hat{j} m$ The electric field strength and its magnitude at a point with radius vector $\vec{r} = 8 \hat{i} - 5 \hat{j} m$ is $(ε_{o} = 8.85 \times 10^{- 12} C^{2} N - m^{- 2})$

Correct answer
A
$4.5 kV m^{- 1}$
B
$45 kV m^{- 1}$
C
$0.45 kV m^{- 1}$
D
$450 kV m^{- 1}$

Solution

Separation vector, $= \vec{r} - \vec{r_{0}}$

$= (8 \hat{i} - 5 \hat{j}) - (2 \hat{i} + 3 \hat{j})$

$= 6 \hat{i} - 8 \hat{j}$

$|\vec{r} - {\vec{r}}_{0}| = \sqrt{{(6)}^{2} + {(- 8)}^{2}} = 10$

Electric field strength,

$E = \frac{1}{4 π ε_{0}} (\frac{q}{r^{2}})$

$= \frac{1}{4 \times π \times 8.85 \times 10^{- 12}} \times \frac{50 \times 10^{- 6}}{100}$

$= 4495.9$

$= 4.495 \times 10^{3} \approx 4.5 kV m^{- 1}$

Hence, the correct answer is option: 'A'.

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A point charge of 50 μC is placed in the XY plane at a location with radius vector r→0=2i^+3j^ m The electric field strength and its magnitude at a point with radius vector r→=8i^-5j^ m is εo=8.85×10-12 C2 N-m-2

Solution

Relevant Questions

A point charge of $50 μ C$ is placed in the $X Y$ plane at a location with radius vector ${\vec{r}}_{0} = 2 \hat{i} + 3 \hat{j} m$ The electric field strength and its magnitude at a point with radius vector $\vec{r} = 8 \hat{i} - 5 \hat{j} m$ is $(ε_{o} = 8.85 \times 10^{- 12} C^{2} N - m^{- 2})$