The figure shows a charge q placed inside a cavity in an uncharged conductor. Now if an external electric field is switched on –

1. only induced charge on the outer surface will redistribute 

2. only induced charge on the inner surface will redistribute 

3. both induced charge on the outer and inner surface will redistribute 

4. force on charge q placed inside the cavity will change

The electric field in a certain region is acting radially outward and is given by \(E=Aa.\) A charge contained in a sphere of radius \(a\) centered at the origin of the field will be given by:

A hollow cylinder has a charge \(q\) coulomb within it (at the geometrical centre). If \(\phi\) is the electric flux in units of Volt-meter associated with the curved surface \(B,\) the flux linked with the plane surface \(A\) in units of volt-meter will be: 
           
1. \(\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)\)
2. \(\frac{q}{2\varepsilon_0}\)
3. \(\frac{\phi}{3}\)
4. \(\frac{q}{\varepsilon_0}-\phi\)

There is a uniform electric field of strength \(10^3\) V/m along the y-axis. A body of mass \(1\) g and charge \(10^{-6}\) C is projected into the field from the origin along the positive x-axis with a velocity \(10\) m/s. Its speed in m/s after \(10\) s is (neglect gravitational effects) 
1. \(10\)
2. \(5\sqrt{2}\)
3. \(10\sqrt{2}\)
4. \(20\)

A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is 

1. 3E along KO 

2. E along OK 

3. E along KO

4. 3E along OK