The Question is from neet-physics electrostatic-charges-and-fields coulombs-law, and it was created by the author VENKATESWARARAO

Question

Two identical small charged spheres, each with mass $m$ and charge $q$, are suspended from a common point by insulating threads of length $L$.  At equilibrium, the angle between the threads is $	heta$. In a different setup with the same parameters, except the charges are now $q$ and $2q$, the angle between the threads becomes $\phi$.  Which of the following relationships holds true if $	heta$ and $\phi$ are small?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Electrostatic force and equilibrium of forces
 **Related Topics**: Coulomb's law, small angle approximation
 **Prerequisites**: Understanding of vector components, equilibrium conditions
## SOLUTION PATHWAY
**Given Information**:
  * Mass of each sphere: $m$
  * Length of threads: $L$
  * Charge of spheres in setup 1: $q, q$
  * Charge of spheres in setup 2: $q, 2q$
  * Angle between threads in setup 1: $	heta$
  * Angle between threads in setup 2: $\phi$
  * $	heta$ and $\phi$ are small angles.
**Method & Steps**:
 1. Analyze the forces acting on one sphere in setup 1.
  Reasoning: The sphere is in equilibrium under the influence of three forces: tension ($T$), gravitational force ($mg$), and electrostatic force ($F_e$).
  Formula/Rule: Coulomb's law: $F_e = \frac{kq^2}{r^2}$, where $k$ is Coulomb's constant and $r$ is the distance between the charges. For small angles, $r \approx x = 2L\sin(	heta/2) \approx L	heta$.
 2. Resolve forces in the horizontal and vertical directions and apply equilibrium conditions.
  Working: In equilibrium, the horizontal component of tension balances the electrostatic force, and the vertical component balances the gravitational force.  For small angles, $T\sin(	heta/2) \approx F_e/2$ and $T\cos(	heta/2) \approx mg$. Thus, $	an(	heta/2) \approx \frac{F_e}{2mg} \approx \frac{kq^2}{2mg(L	heta)^2}$. Since $	heta$ is small, $	an(	heta/2) \approx 	heta/2$. This simplifies to $	heta \approx \sqrt{\frac{2kq^2}{mgL^2}}$.
 3. Repeat for setup 2 and compare.
  Working: For setup 2, the electrostatic force is $F_e' = \frac{k(q)(2q)}{r'^2}$, where $r' \approx L\phi$.  Following the same process, we get $\phi \approx \sqrt{\frac{4kq^2}{mgL^2}} = 2\sqrt{\frac{kq^2}{mgL^2}}$. Comparing with the result for setup 1, we find $\phi \approx 2	heta$.  Since $	heta$ and $\phi$ are small, $	an(\phi) \approx \phi$ and $	an(	heta) \approx 	heta$. Therefore, $	an(\phi) \approx 2	an(	heta)$.
  Result: $	an(\phi) = 2	an(	heta)$
  Verification: The relationship is consistent with the small angle approximation and the dependence of the electrostatic force on the product of the charges.
 ## OPTION ANALYSIS
A] ✓: $	an(\phi) = 2	an(	heta)$ is correct based on the derivation.
B] :x: $\sin(\phi) = 2\sin(	heta)$ is not generally true for small angles.
 C] :x: $\cos(\phi) = \frac{1}{2}\cos(	heta)$ is not generally true for small angles.
D] :x: $\phi = 2	heta$ is approximately true for small angles but the question asks for a relationship that holds true, and this is only approximately true.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly applying Coulomb's law or neglecting the small angle approximation.: Use the appropriate formulas and approximations consistently.
  2.Not resolving forces properly: Clearly resolve forces into horizontal and vertical components for equilibrium analysis.
 **Quick Tips**:
  1.For small angles, $\sin	heta \approx 	an	heta \approx 	heta$ (in radians).
  2.Remember that the electrostatic force is proportional to the product of the charges.

VENKATESWARARAO · Answer

The correct Option is "$	an(\phi) = 2	an(	heta)$"

VENKATESWARARAO · Answer

$\sin(\phi) = 2\sin(	heta)$

VENKATESWARARAO · Answer

$\cos(\phi) = \frac{1}{2}\cos(	heta)$

VENKATESWARARAO · Answer

$\phi = 2	heta$

NEET Physics Electrostatic Charges and Fields Coulomb'S Law MCQs

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