Physics Ii

Chap 16:#6 (II) Two charged dust particles exert a force of 3.2 x  on each other. What will be the force if

they are moved so they are only one-eighth as far apart?

As per Coulomb’s Law

In this case F1= 3.2 x   and r1/r2=8

So final force =

Chap 16: #28 (II) What are the magnitude and direction of the electric field at a point midway between

a -8.0 µC and a +7.0 µC charge 8.0 cm apart? Assume no other charges are nearby.

The distance is same for both charges = 4cm.

Electric field E

The field is directed towards the negative charge. 

Chap 16: #32 (II) The electric field midway between two equal but opposite point charges is 745 N/C, and the distance between the charges is 16.0 cm. What is the magnitude of the charge on each?

 Let each charge =q

So field =

Chap 17: #12 (II) What is the speed of a proton whose kinetic energy is 3.2 keV?

Mass of proton m = 1.67 * 10-27 kg

Now 3.2 keV = 3200 ev = 5.12 * 10-16 J 

KE = mv2/2

v = √(2E/m)

m/s

Chap 17: #20 (II) An electron starts from rest 32.5 cm from a fixed point charge with Q=-0.125 µC. How fast will the electron be moving when it is very far away?

Potential at a distance 32.5 cm is

As we go to infinity the potential due to the charge becomes zero.

So change in potential = 0- (-3.46*103)= 3.46*103 V

So change in energy of the electron =qV = 1.6*10-19 *3.46*103 J.

Now this equals the KE mv2/2.

So speed

Chap 17: #32 (I) A 9500-pF capacitor holds plus and minus charges of 16.5 x  What is the voltage across the capacitor?

Ans: