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This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- electric field at the axis of the ring is E=KQy/ (R²+z²)^3/2 where z is distance .

F=qE=KQqy/ (R²+z²)^3/2

When z<

F= $\frac{2\pi }{w}$ = -Kz

So force is directly proportional to – distance that is completely defined that is follows S.H.M

(b)w= $\pi \sqrt[]{\frac{m}{k}}$

T=2 $\pi \sqrt[]{\frac{m4\pi {\epsilon }_o{d}^3}{2q^2}}$ w=2 $\frac{kqQy}{R3}$

T=2 $\surd \frac{k}{m}$

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Explanation- two charge -q at A and B

AB=AO+OB=2d and x= small distance perpendicular to O

When x

F=qq/4 $\frac{1}{4\pi {\epsilon }_o}\frac{x^2}{\left({10}^{-2}\right)2}$ where AP=BP=r but horizontal components gets cancel out each other and vertical components gets add .

If angle APO=O the net force on q along PO is F’= 2Fcos $\times 10$

= $\pi {\epsilon }_0=8.98755\times {10}^{9}N{m}^2{C}^{-2}$ = $\pi r^2$

When x $\theta$

K= $\frac{2q^2}{4\pi {\epsilon }_o{r}^2}\frac{x}{r}$ ,F $\frac{2q^{2}x}{4\pi {\epsilon }_o{(d^2+x^2)}^{3/2}}$

Force on charge q is proportional to its displacement from the center O and it is directed towards O

Hence we can say that motion of charge would be simple harmonic

Where w= $\frac{2q^{2}x}{4\pi {\epsilon }_o{d}^3}=kx$   and T= $\frac{2q^2}{4\pi {\epsilon }_o{d}^3}$

T= 2 $\propto x$ = 2 $\sqrt[]{\frac{k}{m}}$ = [8π³?_o md³/q²]^1/2  

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This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as we know F= Qq/r²= 1dyne = 1esu of charge²/1cm²

1 esu of charge = M^1/2L^3/2T^—1

• Q=xC where x is dimensionless quantity

 So F= $Q+q_2=q_1$ =1 dyne = 10^-5N

Taking x= 1/3 $Q=q_1-q_2$ ⁹

After solving we get 1/4 $q_1+q_2$

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Some of the best AI & Robotics colleges in Coimbatore are PSG College of Technology, Kumaraguru College of Technology, Sri Krishna College of Engineering and Technology, and Amrita Vishwa Vidyapeetham. These colleges offer good teaching, modern labs, and real-time project exposure. If you are looking for practical, job-focused training, Datamites is a great choice. Datamites provides easy-to-learn AI courses, hands-on projects, expert guidance, and strong placement support, making it perfect for students and working professionals.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- net electric field at plate γ due to two other plate

From plate 1 ,E₁= $\frac{-Q}{S\left(2{\epsilon }_0\right)}totheleft$

From plate 2 ,E₂= $\frac{q}{S\left(2{\epsilon }_0\right)}totheright$

Total electric field E= E₁ + E₂ = $\frac{q-Q}{S\left(2{\epsilon }_0\right)}$ to the left , if Q>q

electric field at o due to plate α = $\frac{-Q}{S\left(2{\epsilon }_0\right)}totheleft$

electric field at o due to plate β = $\frac{q1}{S\left(2{\epsilon }_0\right)}totheright$

electric field at o due to plate γ = $\frac{q2}{S\left(2{\epsilon }_0\right)}totheleft$

as the electric field at o is zero therefore

As there is no loss of charge on collision

Q+q= $\frac{Q+q_2}{S\left(2{\epsilon }_0\right)}=\frac{q_1}{S\left(2{\epsilon }_0\right)}$

On solving these

 q₁= (Q+q/2)= charge on plate β

q₂= (q/2)= charge on plate γ

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Most AI & Robotics colleges in Coimbatore admit students through merit-based scores; however, students may also be admitted through the entrance exams listed below:

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Of course! Students can go through the table below for some details about the AI & Robotics colleges in Coimbatore:

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There are 20+ AI and Robotics colleges in Coimbatore where students can pursue the degree at various course levels like certificates, diplomas, BSc, B.E./B.Tech, and M.E./M.Tech, etc.

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A student who aspires to build a career in the field of Computer Science and IT should definitely pursue a BCA course. It will equip them with the necessary skills and certifications to pursue a career in this field. 

Students can evaluate the monetary benefits of pursuing a BCA at West Bengal colleges by comparing the course fees and the placement salary at these colleges:

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