The three charges q/2, q and q/2 are placed at the corners A, B and C of a square of side 'a' as shown in figure. The magnitude of electric field (E) at the corner D of the square, is :

q \over 4\pi \in _0a^2( 1 \over \sqrt 2 + 1 \over 2 )

q \over 4\pi \in _0a^2( 1 + 1 \over \sqrt 2 )

q \over 4\pi \in _0a^2( 1 - 1 \over \sqrt 2 )

q \over 4\pi \in _0a^2( 1 \over \sqrt 2 - 1 \over 2 )

Electrostatics PYQ — JEE Main 2022 · 28 Jun · Shift 1 · Q51 with Solution

View written solutionFree

Correct answer: A

Set up the square and charges

Let the square be $ABCD$ with side $a$ , and charges placed at:

$A : \dfrac{q}{2}$
$B : q$
$C : \dfrac{q}{2}$

We need the electric field at corner $D$ .

Take coordinates:

$A(0,a)$
$B(a,a)$
$C(a,0)$
$D(0,0)$

Electric field at $D$ due to charge at $A$

Distance $AD=a$ .

So,

E_A = \frac{1}{4\pi\varepsilon_0}\frac{q/2}{a^2} = \frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12

Direction: from $A$ to $D$ , i.e. vertically downward.

Hence,

\vec E_A = -\frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12\,\hat j

Electric field at $D$ due to charge at $C$

Distance $CD=a$ .

So,

E_C = \frac{1}{4\pi\varepsilon_0}\frac{q/2}{a^2} = \frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12

Direction: from $C$ to $D$ , i.e. horizontally left.

Hence,

\vec E_C = -\frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12\,\hat i

Electric field at $D$ due to charge at $B$

Distance $BD=\sqrt{a^2+a^2}=a\sqrt2$ .

So,

E_B = \frac{1}{4\pi\varepsilon_0}\frac{q}{(a\sqrt2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{q}{2a^2} = \frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12

Direction is along diagonal $BD$ from $B$ to $D$ , i.e. equally left and downward.

Therefore its components are:

E_{Bx}=E_{By}=E_B\cos45^\circ = \frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12\cdot \frac{1}{\sqrt2}

with negative signs in both $x$ and $y$ directions:

\vec E_B = -\frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac{1}{2\sqrt2}\,\hat i -\frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac{1}{2\sqrt2}\,\hat j

Add components

Total $x$ -component:

E_x = -\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac12+\frac{1}{2\sqrt2}\right)

Total $y$ -component:

E_y = -\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac12+\frac{1}{2\sqrt2}\right)

So the two components are equal.

Magnitude of resultant electric field

Let

K=\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac12+\frac{1}{2\sqrt2}\right)

Then

E=\sqrt{E_x^2+E_y^2}=\sqrt{K^2+K^2}=\sqrt2\,K

Thus,

E=\sqrt2\cdot \frac{q}{4\pi\varepsilon_0 a^2}\left(\frac12+\frac{1}{2\sqrt2}\right)

Simplify:

E=\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac{\sqrt2}{2}+\frac12\right)

Since $\dfrac{\sqrt2}{2}=\dfrac{1}{\sqrt2}$ ,

E=\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac{1}{\sqrt2}+\frac12\right)

Compare with options

This matches:

\boxed{\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac{1}{\sqrt2}+\frac12\right)}

which is Option A.

Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.

More from Electrostatics

View written solutionFree

Correct answer: A

Set up the square and charges

Let the square be $ABCD$ with side $a$ , and charges placed at:

$A : \dfrac{q}{2}$
$B : q$
$C : \dfrac{q}{2}$

We need the electric field at corner $D$ .

Take coordinates:

$A(0,a)$
$B(a,a)$
$C(a,0)$
$D(0,0)$

Electric field at $D$ due to charge at $A$

Distance $AD=a$ .

So,

E_A = \frac{1}{4\pi\varepsilon_0}\frac{q/2}{a^2} = \frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12

Direction: from $A$ to $D$ , i.e. vertically downward.

Hence,

\vec E_A = -\frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12\,\hat j

Electric field at $D$ due to charge at $C$

Distance $CD=a$ .

So,

E_C = \frac{1}{4\pi\varepsilon_0}\frac{q/2}{a^2} = \frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12

Direction: from $C$ to $D$ , i.e. horizontally left.

Hence,

\vec E_C = -\frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12\,\hat i

Electric field at $D$ due to charge at $B$

Distance $BD=\sqrt{a^2+a^2}=a\sqrt2$ .

So,

E_B = \frac{1}{4\pi\varepsilon_0}\frac{q}{(a\sqrt2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{q}{2a^2} = \frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12

Direction is along diagonal $BD$ from $B$ to $D$ , i.e. equally left and downward.

Therefore its components are:

E_{Bx}=E_{By}=E_B\cos45^\circ = \frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac12\cdot \frac{1}{\sqrt2}

with negative signs in both $x$ and $y$ directions:

\vec E_B = -\frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac{1}{2\sqrt2}\,\hat i -\frac{q}{4\pi\varepsilon_0 a^2}\cdot \frac{1}{2\sqrt2}\,\hat j

Add components

Total $x$ -component:

E_x = -\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac12+\frac{1}{2\sqrt2}\right)

Total $y$ -component:

E_y = -\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac12+\frac{1}{2\sqrt2}\right)

So the two components are equal.

Magnitude of resultant electric field

Let

K=\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac12+\frac{1}{2\sqrt2}\right)

Then

E=\sqrt{E_x^2+E_y^2}=\sqrt{K^2+K^2}=\sqrt2\,K

Thus,

E=\sqrt2\cdot \frac{q}{4\pi\varepsilon_0 a^2}\left(\frac12+\frac{1}{2\sqrt2}\right)

Simplify:

E=\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac{\sqrt2}{2}+\frac12\right)

Since $\dfrac{\sqrt2}{2}=\dfrac{1}{\sqrt2}$ ,

E=\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac{1}{\sqrt2}+\frac12\right)

Compare with options

This matches:

\boxed{\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac{1}{\sqrt2}+\frac12\right)}

which is Option A.

Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.

Electrostatics question

Electrostatics question

More from Electrostatics

Electrostatics question

Electrostatics question

More from Electrostatics