Six charges are placed around a regular hexagon of side length a as shown in the figure. Five of them have charge q , and the remaining one has charge x . The perpendicular from each charge to the nearest hexagon side passes through the center 0 of the hexagon and is bisected by the side. Which of the following statement(s) is(are) correct in SI units?

When x=q , the magnitude of the electric field at 0 is zero., When x=-q , the magnitude of the electric field at 0 is \fracq6 \pi \epsilon_o a^2 ., When x=2 q , the potential at 0 is \frac7 q4 \sqrt3 \pi \epsilon_o a .

When x=q , the magnitude of the electric field at 0 is zero.

When x=-q , the magnitude of the electric field at 0 is \fracq6 \pi \epsilon_o a^2 .

When x=2 q , the potential at 0 is \frac7 q4 \sqrt3 \pi \epsilon_o a .

When x=-3 q , the potential at 0 is -\frac3 q4 \sqrt3 \pi \epsilon_o a .

Electrostatics PYQ — JEE Advanced 2022 · Shift 1 · Q49 with Solution

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Correct answer: A, B, C

Geometry of the arrangement

A regular hexagon has side length $a$ . For a regular hexagon,

circumradius $R = a$
apothem $r = \dfrac{\sqrt{3}}{2}a$

The statement says that each charge lies on the perpendicular to a side through the center, and that side bisects this perpendicular segment. Hence each charge is the reflection of the center across a side, so the distance of each charge from the center is

$2r = 2\cdot \frac{\sqrt{3}}{2}a = \sqrt{3}a.$

So every charge is at the same distance

$d=\sqrt{3}a$

from the center $O$ .

Electric potential at the center

Potential is a scalar, so it simply adds:

$V_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{5q+x}{d}$

with $d=\sqrt{3}a$ . Therefore,

$V_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{5q+x}{\sqrt{3}a}.$

Electric field at the center

The six charge positions are equally spaced in angle around the center. If all six charges were $q$ , the net electric field at the center would be zero by symmetry.

Now five charges are $q$ and one is $x$ . Think of this as:

six charges each of value $q$ , plus
an extra charge $(x-q)$ at the position where the charge is $x$ .

Since the field due to six equal charges $q$ cancels out,

$\vec E_O = \text{field due only to }(x-q)\text{ at that one position.}$

Magnitude of field at center due to a charge of magnitude $|x-q|$ at distance $d$ is

$E_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{|x-q|}{d^2}.$

Since

$d^2=(\sqrt{3}a)^2=3a^2,$

we get

$E_O = \frac{|x-q|}{12\pi\varepsilon_0 a^2}.$

Check each option

Option A: When $x=q$ , the magnitude of electric field at $O$ is zero.

Using

$E_O = \frac{|x-q|}{12\pi\varepsilon_0 a^2},$

if $x=q$ ,

$E_O = 0.$

So A is correct.

Option B: When $x=-q$ , the magnitude of electric field at $O$ is $\dfrac{q}{6\pi\varepsilon_0 a^2}$ .

Substitute $x=-q$ :

$|x-q|=|-q-q|=2q.$

Hence

$E_O=\frac{2q}{12\pi\varepsilon_0 a^2}=\frac{q}{6\pi\varepsilon_0 a^2}.$

So B is correct.

Option C: When $x=2q$ , the potential at $O$ is $\dfrac{7q}{4\sqrt{3}\pi\varepsilon_0 a}$ .

Using

$V_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{5q+x}{\sqrt{3}a},$

for $x=2q$ ,

= \frac{7q}{4\sqrt{3}\pi\varepsilon_0 a}.$$ So **C is correct**. --- ### Option D: When $x=-3q$, the potential at $O$ is $-\dfrac{3q}{4\sqrt{3}\pi\varepsilon_0 a}$. For $x=-3q$, $$5q+x=5q-3q=2q.$$ Thus $$V_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{2q}{\sqrt{3}a} = \frac{q}{2\sqrt{3}\pi\varepsilon_0 a},$$ which is **positive**, not negative. So **D is incorrect**. --- 5. **Final answer** The correct options are: $$\boxed{A,\ B,\ C}$$ This matches the stored correct answer.

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Correct answer: A, B, C

Geometry of the arrangement

A regular hexagon has side length $a$ . For a regular hexagon,

circumradius $R = a$
apothem $r = \dfrac{\sqrt{3}}{2}a$

$2r = 2\cdot \frac{\sqrt{3}}{2}a = \sqrt{3}a.$

So every charge is at the same distance

$d=\sqrt{3}a$

from the center $O$ .

Electric potential at the center

Potential is a scalar, so it simply adds:

$V_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{5q+x}{d}$

with $d=\sqrt{3}a$ . Therefore,

$V_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{5q+x}{\sqrt{3}a}.$

Electric field at the center

The six charge positions are equally spaced in angle around the center. If all six charges were $q$ , the net electric field at the center would be zero by symmetry.

Now five charges are $q$ and one is $x$ . Think of this as:

six charges each of value $q$ , plus
an extra charge $(x-q)$ at the position where the charge is $x$ .

Since the field due to six equal charges $q$ cancels out,

$\vec E_O = \text{field due only to }(x-q)\text{ at that one position.}$

Magnitude of field at center due to a charge of magnitude $|x-q|$ at distance $d$ is

$E_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{|x-q|}{d^2}.$

Since

$d^2=(\sqrt{3}a)^2=3a^2,$

we get

$E_O = \frac{|x-q|}{12\pi\varepsilon_0 a^2}.$

Check each option

Option A: When $x=q$ , the magnitude of electric field at $O$ is zero.

Using

$E_O = \frac{|x-q|}{12\pi\varepsilon_0 a^2},$

if $x=q$ ,

$E_O = 0.$

So A is correct.

Option B: When $x=-q$ , the magnitude of electric field at $O$ is $\dfrac{q}{6\pi\varepsilon_0 a^2}$ .

Substitute $x=-q$ :

$|x-q|=|-q-q|=2q.$

Hence

$E_O=\frac{2q}{12\pi\varepsilon_0 a^2}=\frac{q}{6\pi\varepsilon_0 a^2}.$

So B is correct.

Option C: When $x=2q$ , the potential at $O$ is $\dfrac{7q}{4\sqrt{3}\pi\varepsilon_0 a}$ .

Using

$V_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{5q+x}{\sqrt{3}a},$

for $x=2q$ ,

= \frac{7q}{4\sqrt{3}\pi\varepsilon_0 a}.$$ So **C is correct**. --- ### Option D: When $x=-3q$, the potential at $O$ is $-\dfrac{3q}{4\sqrt{3}\pi\varepsilon_0 a}$. For $x=-3q$, $$5q+x=5q-3q=2q.$$ Thus $$V_O = \frac{1}{4\pi\varepsilon_0}\cdot \frac{2q}{\sqrt{3}a} = \frac{q}{2\sqrt{3}\pi\varepsilon_0 a},$$ which is **positive**, not negative. So **D is incorrect**. --- 5. **Final answer** The correct options are: $$\boxed{A,\ B,\ C}$$ This matches the stored correct answer.

Electrostatics question

Electrostatics question

Option A: When $x=q$ , the magnitude of electric field at $O$ is zero.

Option B: When $x=-q$ , the magnitude of electric field at $O$ is $\dfrac{q}{6\pi\varepsilon_0 a^2}$ .

Option C: When $x=2q$ , the potential at $O$ is $\dfrac{7q}{4\sqrt{3}\pi\varepsilon_0 a}$ .

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Electrostatics question

Electrostatics question

Option A: When $x=q$ , the magnitude of electric field at $O$ is zero.

Option B: When $x=-q$ , the magnitude of electric field at $O$ is $\dfrac{q}{6\pi\varepsilon_0 a^2}$ .

Option C: When $x=2q$ , the potential at $O$ is $\dfrac{7q}{4\sqrt{3}\pi\varepsilon_0 a}$ .

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Option A: When x=qx=qx=q, the magnitude of electric field at OOO is zero.

Option B: When x=−qx=-qx=−q, the magnitude of electric field at OOO is q6πε0a2\dfrac{q}{6\pi\varepsilon_0 a^2}6πε0​a2q​.

Option C: When x=2qx=2qx=2q, the potential at OOO is 7q43πε0a\dfrac{7q}{4\sqrt{3}\pi\varepsilon_0 a}43​πε0​a7q​.

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Option A: When x=qx=qx=q, the magnitude of electric field at OOO is zero.

Option B: When x=−qx=-qx=−q, the magnitude of electric field at OOO is q6πε0a2\dfrac{q}{6\pi\varepsilon_0 a^2}6πε0​a2q​.

Option C: When x=2qx=2qx=2q, the potential at OOO is 7q43πε0a\dfrac{7q}{4\sqrt{3}\pi\varepsilon_0 a}43​πε0​a7q​.

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Option A: When $x=q$ , the magnitude of electric field at $O$ is zero.

Option B: When $x=-q$ , the magnitude of electric field at $O$ is $\dfrac{q}{6\pi\varepsilon_0 a^2}$ .

Option C: When $x=2q$ , the potential at $O$ is $\dfrac{7q}{4\sqrt{3}\pi\varepsilon_0 a}$ .

Option A: When $x=q$ , the magnitude of electric field at $O$ is zero.

Option B: When $x=-q$ , the magnitude of electric field at $O$ is $\dfrac{q}{6\pi\varepsilon_0 a^2}$ .

Option C: When $x=2q$ , the potential at $O$ is $\dfrac{7q}{4\sqrt{3}\pi\varepsilon_0 a}$ .