Two point charges $q_{1}(\sqrt{10} \mu \mathrm{C})$ and $q_{2}(-25 \mu \mathrm{C})$ are placed on the $x$-axis at $x=1 \mathrm{~m}$ and $x=4 \mathrm{~m}$ respectively. The electric field (in $\mathrm{V} / \mathrm{m}$) at a point $y=3 \mathrm{~m}$ on $y$-axis is, $\left[\right.$ take $\left.\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\right]$

(A) $(63 \hat{i}-27 \hat{j}) \times 10^{2}$

(B) $(-63 \hat{i}+27 \hat{j}) \times 10^{2}$

(C) $(81 \hat{\mathrm{i}}-81 \hat{\mathrm{j}}) \times 10^{2}$

(D) $(-81 \hat{i}+81 \hat{j}) \times 10^{2}$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Electrostatics

Two point charges

q_{1} (10 μ C)

and

q_{2} (- 25 μ C)

are placed on the

x

-axis at

x = 1 m

and

x = 4 m

respectively. The electric field (in

V / m

) at a point

y = 3 m

y

-axis is,

[

take

\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} Nm^{2} C^{- 2}]

Answer: (A)

Find each charge's field at the point

(0, 3)

and add as vectors. From

q_{1} = + 10 μ C

(1, 0)

the vector to the field point is

- \hat{i} + 3 \hat{j}

with length

10

; from

q_{2} = - 25 μ C

(4, 0)

it is

- 4 \hat{i} + 3 \hat{j}

with length

5

. Using

E = \frac{k q}{r ^{3}} r

for each, the first contributes

900 (- \hat{i} + 3 \hat{j})

and the second

- 1800 (- 4 \hat{i} + 3 \hat{j})

(field in V/m). Adding,

E = (63 \hat{i} - 27 \hat{j}) \times 1 0^{2} V/m

. The listed choice prints the second unit vector as

\hat{i}

, but it should read

\hat{j}

-- a typo in the paper.

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option