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Electrostatics question

2019 · 12 Apr · Shift 1 · Q55
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Electrostatics question

2019 · 12 Apr · Shift 1 · Q55

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A point dipole p→=−p0x^\overrightarrow p = - {p_0}\widehat xp​=−p0​x is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively: (Take V= 0 at infinity)
  1. A
    ∣p→∣4π∈0d2,−p→4π∈0d3{{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}4π∈0​d2​p​​​,4π∈0​d3−p​​
  2. B
    0,p→4π∈0d30,{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}0,4π∈0​d3p​​
  3. C
    ∣p→∣4π∈0d2,p→4π∈0d3{{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}4π∈0​d2​p​​​,4π∈0​d3p​​
  4. D
    0,−p→4π∈0d30,{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}0,4π∈0​d3−p​​
View written solutionFree

Correct answer: D

  1. Given dipole and observation point

A point dipole is placed at the origin:

p⃗=−p0x^\vec p = -p_0\hat xp​=−p0​x^

We need the potential and electric field at a point on the y-axis at distance ddd, i.e. at

r⃗=dy^\vec r = d\hat yr=dy^​

with

r=d,r^=y^r=d, \qquad \hat r = \hat yr=d,r^=y^​
  1. Potential due to a point dipole

The potential due to a dipole at position r⃗\vec rr is

V(r⃗)=14πε0p⃗⋅r^r2V(\vec r)=\frac{1}{4\pi\varepsilon_0}\frac{\vec p\cdot \hat r}{r^2}V(r)=4πε0​1​r2p​⋅r^​

Here,

p⃗⋅r^=(−p0x^)⋅y^=0\vec p\cdot \hat r = (-p_0\hat x)\cdot \hat y = 0p​⋅r^=(−p0​x^)⋅y^​=0

Therefore,

V=14πε00d2=0V=\frac{1}{4\pi\varepsilon_0}\frac{0}{d^2}=0V=4πε0​1​d20​=0

So the potential on the y-axis is

V=0V=0V=0
  1. Electric field due to a point dipole

The electric field of a point dipole is

E⃗=14πε0r3[3(p⃗⋅r^)r^−p⃗]\vec E = \frac{1}{4\pi\varepsilon_0 r^3}\left[3(\vec p\cdot \hat r)\hat r-\vec p\right]E=4πε0​r31​[3(p​⋅r^)r^−p​]

Since p⃗⋅r^=0\vec p\cdot \hat r=0p​⋅r^=0 on the y-axis,

E⃗=14πε0d3(0−p⃗)=−p⃗4πε0d3\vec E = \frac{1}{4\pi\varepsilon_0 d^3}(0-\vec p) = -\frac{\vec p}{4\pi\varepsilon_0 d^3}E=4πε0​d31​(0−p​)=−4πε0​d3p​​

Now substitute p⃗=−p0x^\vec p=-p_0\hat xp​=−p0​x^:

E⃗=−−p0x^4πε0d3=p0x^4πε0d3\vec E = -\frac{-p_0\hat x}{4\pi\varepsilon_0 d^3} = \frac{p_0\hat x}{4\pi\varepsilon_0 d^3}E=−4πε0​d3−p0​x^​=4πε0​d3p0​x^​

Thus, in vector form,

E⃗=−p⃗4πε0d3\vec E = -\frac{\vec p}{4\pi\varepsilon_0 d^3}E=−4πε0​d3p​​
  1. Match with options

We found:

  • Potential: 000
  • Electric field: −p⃗4πε0d3-\dfrac{\vec p}{4\pi\varepsilon_0 d^3}−4πε0​d3p​​

This matches Option D.

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So, the stored answer is correct.

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