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Electrostatics question

2021 · 24 Feb · Shift 1 · Q45
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Electrostatics question

2021 · 24 Feb · Shift 1 · Q45

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is −-− Q. The electric field at the centre of cube is : JEE Main 2021 (Online) 24th February Morning Shift Physics - Electrostatics Question 162 English
  1. A
    2Q33πε0a2(x^+y^+z^){{2Q} \over {3\sqrt 3 \pi {\varepsilon _0}{a^2}}}\left( {\widehat x + \widehat y + \widehat z} \right)33​πε0​a22Q​(x+y​+z)
  2. B
    −Q33πε0a2(x^+y^+z^){{ - Q} \over {3\sqrt 3 \pi {\varepsilon _0}{a^2}}}\left( {\widehat x + \widehat y + \widehat z} \right)33​πε0​a2−Q​(x+y​+z)
  3. C
    Q33πε0a2(x^+y^+z^){Q \over {3\sqrt 3 \pi {\varepsilon _0}{a^2}}}\left( {\widehat x + \widehat y + \widehat z} \right)33​πε0​a2Q​(x+y​+z)
  4. D
    −2Q33πε0a2(x^+y^+z^){{ - 2Q} \over {3\sqrt 3 \pi {\varepsilon _0}{a^2}}}\left( {\widehat x + \widehat y + \widehat z} \right)33​πε0​a2−2Q​(x+y​+z)
View written solutionFree

Correct answer: D

  1. Set up the cube coordinates

Let the cube have vertices at

(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a), (a,a,a)(0,0,0),\ (a,0,0),\ (0,a,0),\ (0,0,a),\ (a,a,0),\ (a,0,a),\ (0,a,a),\ (a,a,a)(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a), (a,a,a)

with the origin carrying charge −Q-Q−Q and all other 7 vertices carrying charge +Q+Q+Q.

The centre of the cube is at

(a2,a2,a2).\left(\frac a2,\frac a2,\frac a2\right).(2a​,2a​,2a​).
  1. Use symmetry cleverly

If all 8 vertices had charge +Q+Q+Q, then by symmetry the electric field at the centre would be zero:

E⃗all +Q=0.\vec E_{\text{all }+Q}=0.Eall +Q​=0.

But in the actual problem, the charge at the origin is changed from +Q+Q+Q to −Q-Q−Q.

So compared to the symmetric case, the net change is equivalent to adding an extra charge −2Q-2Q−2Q at the origin:

(+Q→−Q)⇒change=−2Q.(+Q \to -Q) \Rightarrow \text{change} = -2Q.(+Q→−Q)⇒change=−2Q.

Hence, the required electric field at the centre is just the field due to a point charge −2Q-2Q−2Q placed at the origin.

  1. Distance from origin to centre

The distance is

r=(a2)2+(a2)2+(a2)2=a32.r=\sqrt{\left(\frac a2\right)^2+\left(\frac a2\right)^2+\left(\frac a2\right)^2} =\frac{a\sqrt3}{2}.r=(2a​)2+(2a​)2+(2a​)2​=2a3​​.

So

r2=3a24.r^2=\frac{3a^2}{4}.r2=43a2​.
  1. Magnitude of electric field due to charge −2Q-2Q−2Q

Using

E=14πε0∣q∣r2,E=\frac{1}{4\pi\varepsilon_0}\frac{|q|}{r^2},E=4πε0​1​r2∣q∣​,

we get

E=14πε0⋅2Q3a2/4=14πε0⋅8Q3a2=2Q3πε0a2.E=\frac{1}{4\pi\varepsilon_0}\cdot \frac{2Q}{3a^2/4} =\frac{1}{4\pi\varepsilon_0}\cdot \frac{8Q}{3a^2} =\frac{2Q}{3\pi\varepsilon_0 a^2}.E=4πε0​1​⋅3a2/42Q​=4πε0​1​⋅3a28Q​=3πε0​a22Q​.
  1. Direction of the field

Since the charge is negative, the electric field at the centre points towards the origin.

The vector from origin to centre is proportional to

x^+y^+z^.\hat x+\hat y+\hat z.x^+y^​+z^.

Therefore, the field towards the origin is along

−(x^+y^+z^).-(\hat x+\hat y+\hat z).−(x^+y^​+z^).

The unit vector in the body-diagonal direction is

x^+y^+z^3.\frac{\hat x+\hat y+\hat z}{\sqrt3}.3​x^+y^​+z^​.

Hence,

E⃗=−2Q3πε0a2⋅x^+y^+z^3=−2Q33πε0a2(x^+y^+z^).\vec E=-\frac{2Q}{3\pi\varepsilon_0 a^2}\cdot \frac{\hat x+\hat y+\hat z}{\sqrt3} = -\frac{2Q}{3\sqrt3\pi\varepsilon_0 a^2}(\hat x+\hat y+\hat z).E=−3πε0​a22Q​⋅3​x^+y^​+z^​=−33​πε0​a22Q​(x^+y^​+z^).
  1. Match with options

This matches:

D: −2Q33πε0a2(x^+y^+z^)\boxed{\text{D: }-\frac{2Q}{3\sqrt3\pi\varepsilon_0 a^2}(\hat x+\hat y+\hat z)}D: −33​πε0​a22Q​(x^+y^​+z^)​
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