Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field. Reason (R): In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions. In the light of the above statements, choose the most appropriate answer from the options given below:

Solution & Explanation

### Related Formula vecP_textnet = sum vecp_i where: vecP_textnet = net dipole moment of the dielectric vecp_i = dipole moment of the individual i-th molecule ### Core Logic No external electric field is present (E_textext = 0). Due to thermal agitation, all molecular permanent dipoles are randomly oriented in space: vecP_textnet = 0 quad textwhen vecE_textext = 0 Thus: 1. Assertion (A) is false because it claims the net dipole moment is non-zero even without an external field. 2. Reason (R) is true because it correctly describes that different permanent dipoles are randomly oriented. ### Step 1: Final Conclusion Therefore, (A) is not correct but (R) is correct. ### Pattern Recognition Sees: "polar dielectric" + "no external field" → net bulk dipole moment is always zero. Trap: Confusing the molecular level with the macroscopic level. Each molecule in a polar dielectric has a permanent dipole moment, but the macro substance has zero net moment due to random thermal orientations. Shortcut: No external field means vectors cancel globally, which implies zero net moment. Thus (A) is false immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

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Q42 jee_main_2024_31_jan_morning Electric Field Zero Point
Two charges q and 3q are separated by a distance 'r' in air. At a distance x from charge q, the resultant electric field is zero. The value of x is :
  • A. frac(1 + sqrt3)r
  • B. fracr3(1 + sqrt3)
  • C. fracr(1 + sqrt3)
  • D. r(1 + sqrt3)

Solution

### Related Formula E = frackqx^2 ### Core Logic
Electric Field Zero Point diagram for Q42 - JEE Main 2024 Morning
Electric Field Zero Point diagram for Q42 - JEE Main 2024 Morning
For the net electric field to be zero at point P situated at distance x from charge q, the electric fields produced by both charges must be equal in magnitude and opposite in direction. Let the charges be placed at ends of a line. Point P is between them since both charges are of the same sign. (vecE_textnet)_P = 0 frackqx^2 = frack(3q)(r-x)^2 ### Step 2: Solving for x Taking square roots on both sides: frac1x = fracsqrt3r-x r - x = sqrt3x r = x(sqrt3 + 1) x = fracrsqrt3 + 1 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics
Q52 jee_main_2024_31_jan_morning Capacitance With Dielectric
A parallel plate capacitor with plate separation 5 mathrm~mm is charged up by a battery. It is found that on introducing a dielectric sheet of thickness 2 mathrm~mm, while keeping the battery connections intact, the capacitor draws 25 \% more charge from the battery than before. The dielectric constant of the sheet is _____.
Numerical Answer. Answer: 2 to 2

Solution

### Related Formula C = fracvarepsilon_0 Ad C' = fracvarepsilon_0 Ad - t + fractK Q = CV ### Core Logic Initially, the charge stored without the dielectric is: Q_i = fracA varepsilon_0d V After introducing a dielectric of thickness t, the new capacitance C' leads to a new charge Q_f: Q_f = fracA varepsilon_0 Vd - t + fractK ### Step 2: Charge Relationship Given that the capacitor draws 25\% more charge: Q_f = 1.25 Q_i = frac54 Q_i Equating the expressions: fracA varepsilon_0 Vd - t + fractK = 1.25 left( fracA varepsilon_0 Vd right) frac15 - 2 + frac2K = frac1.255 frac13 + frac2K = frac1.255 = frac14 3 + frac2K = 4 frac2K = 1 Rightarrow K = 2 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

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