Electric charges and fields
Electric Charges and Fields
Problems
Question1. Three-point
charges of +2 μC, -3 μC, and -3 μC are kept at the vertices A, B, and C
respectively of an equilateral triangle of the side of 20 cm as shown in the figure
(a). What should be the sign and magnitude of the charge to be placed at the
mid-point (M) of side BC so that the charge at A remains in equilibrium?
Question2. Calculate the
electric field strength required to just support a water drop of mass 10-3
kg and having a charge of 1.6 × 10-19 C.
Solution. Here m = 10-3 kg,
q = 1.6 × 10-19 C
Let E be the strength of the electric
field required to just support the water drop. Then
Force
on water drop due to electric field = Weight of water drop
Or
qE = mg
Question3. Calculate the
voltage needed to balance an oil drop carrying 10 electrons when located
between the plates of a capacitor which are 5 mm apart. The mass of oil drop is
3 × 10-16 kg. Take g = 10 ms-2.
Question4. How many
electrons should be removed from a coin of mass 1.6 g, so that it may just
float in an electric field of intensity 109 NC-1,
directed upward?
Solution. Here m = 1.6 g = 1.6 × 10-3 kg,
E = 109 NC-1
Let n be the number of electrons
removed from the coin,
Then charge on the coin,
q = + ne
When the coin just floats,
Upward force of electric field = Weight of coin
Or qE or neE = mg
Question5. A pendulum of
mass 80 milligrams carrying a charge of 2 × 10-8 C is at rest in a horizontal uniform electric field of 2 × 104 Vm-1. Find
the tension in the thread of the pendulum and the angle it makes with the
vertical.
Solution. Here m = 80 mg = 80 × 10-6 kg
q = 2 ×
10-8 C, E = 2 × 104 Vm-1
Question6. An electron
moves a distance of 6 cm when accelerated from rest by an electric field of
strength 2 × 104 NC-1. Calculate the time of travel. The
mass and charge of the electron are 9 × 10-19 kg and 1.6 × 10-19
C respectively.
Solution. Force exerted on the
electron by the electric field,
Question7. An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 NC-1. Given in figure (a), the direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance. Given in figure (b), compute the time of falls in each case. Contrast the situation (a) with that of ‘free fall under the gravity’.
Thus the heavier particle takes a
greater time to fall through the same distance. This is in contrast to the
situation of ‘free fall under gravity’ where the time of fall is independent of
the mass of the body. Here the acceleration due to gravity ‘g’ being negligibly
small has been ignored.
Question10. An electric
field E is set up between the two parallel plates of a capacitor, as shown in
the figure, an electron enters the field symmetrically between the plates with a
speed v0. The length of each plate l. find the angle of deviation of the path of the electron as it
comes out of the field.
Comments
Post a Comment
please do not enter any spam link in the comment box