ELECTROSTATICS (PART -1) - NEET MCQ

 By symmetry of problem the components of force on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will be along CO. Under the action of this force charge Q will move towards O. If at any time charge Q is at a distance x from O. Net force on charge Q  \[{{F}_{net}}\Rightarrow 2F\cos \theta =2\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{-qQ}{({{a}^{2}}+{{x}^{2}})}\times \frac{x}{{{({{a}^{2}}+{{x}^{2}})}^{{1}/{{}}\;2}}}\] i.e., \[{{F}_{net}}=-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2qQx}{{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{{3}/{{}}\;2}}}\] As the restoring force Fnet is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmoni

Charge will move along the circular line of force because 

x2+y2=1

 is the equation of circle in xy-plane.

• Because of the presence of positive test charge q0 in front of positively charged ball, charge on the ball will be redistributed, less charge on the front half surface and more charge on the back half surface. As a result of this net force F between ball and point charge will decrease i.e. actual electric field will be greater than 

  F/q0.

Electric field at a distance R is only due to sphere because electric field due to shell inside it is always zero. Hence electric field =

14πε0.3QR2

Initially , when the charge is at rest then 

initial K_i=0

when the particle aquires some velocity

then,K_f>0

Now, according to work energy theorem

W=k_f-Ki

K_f=force x displacement

or,k(f)=(qE)(y)

=qEy