An electron enters with kinetic energy `KE_1`, between the plates of a capacitor with a velocity making angle `alpha` as shown and leaves with `KE_2` making angle `beta`?. Find `(KE_1)/(KE_2)` ?

IF an electorn enters into a space between the plates of a parallel plate capacitor at an an angle alpha with the plates an leaves at an angle beta to the plates find the ratio of its kinetic energy while entering the capacitor of that while leaving.

If an electron enters a space between the plates of a parallel plate capacitor at an angle theta_(1) with the plates and leaves at an angle theta_(2) to the plates, the ratio of its kinetic energy while entering the capacitor to that while leaving will be

when a positively charged particle is projected with a velocity v_0 in the region between the horizontal plates of the capacitor then it leaves by making an angle theta with the horizontal , as shown. If theta=theta_1 is for proton and theta=theta_2 for an alpha - particle then the value of (tan theta_1)/(tantheta_2)

A projectile is fired with kinetic energy 1 k J. If the range is maximum , what is its K.E. at the highest point ?

Potential energy (PE_(n)) and kinetic energy (KE_(n)) of electron in nth orbit are related as

The ratio of KE of a planet at the points 1 and 2 is :

An electron enters into the space of uniform field of finite width making an angle 60^(@) to the field lines and leaves at angle 30^(@) to the field lines. The ratio of its kinetic energy while entering the electric field to the while leave will be :

An electron is 2000 times lighter than a proton. An electron and a proton are moving with such a velocity that de-Broglie wave associated with them is 1Å. The ratio of their K.E. will be :