Advertisements
Advertisements
Question
Two particles A and B, each carrying a charge Q, are held fixed with a separation dbetween them. A particle C of mass m and charge q is kept at the middle point of the line AB. If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it?
Solution
The charge q is displaced by a distance x on the perpendicular bisector of AB.
As shown in the figure, the horizontal component of the force is balanced.
\[\sin\theta = \frac{x}{\sqrt{\left( \frac{d}{2} \right)^2 + x^2}}\]
Total vertical component of the force,
\[F' = 2F\sin\theta\]
\[F' = 2 \times \frac{1}{4\pi \epsilon_0} \times \frac{qQ}{\left( \frac{d}{2} \right)^2 + x^2} \times \frac{x}{\sqrt{\left( \frac{d}{2} \right)^2 + x^2}}\]
\[ \Rightarrow F' = \frac{1}{2\pi \epsilon_0} \times \frac{qQx}{\left[ \left( \frac{d}{2} \right)^2 + x^2 \right]^{3/2}}\]
This is the net electric force experienced by the charge q.
APPEARS IN
RELATED QUESTIONS
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N.
- What is the distance between the two spheres?
- What is the force on the second sphere due to the first?
Suppose that the particle is an electron projected with velocity vx = 2.0 × 106 m s−1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (|e| = 1.6 × 10−19 C, me = 9.1 × 10−31 kg)
Find the dimensional formula of ε0.
Two charged particles are placed 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?
Suppose an attractive nuclear force acts between two protons which may be written as F=Ce−kr/r2. Suppose that k = 1 fermi−1 and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.
Four equal charges of 2.0 × 10−6 C each are fixed at the four corners of a square of side 5 cm. Find the Coulomb's force experienced by one of the charges due to the other three.
Two small spheres, each with a mass of 20 g, are suspended from a common point by two insulating strings of length 40 cm each. The spheres are identically charged and the separation between the balls at equilibrium is found to be 4 cm. Find the charge on each sphere.
Two particles A and B, each carrying a charge Q, are held fixed with a separation dbetween them. A particle C of mass m and charge q is kept at the middle point of the line AB. Assuming x<<d, show that this force is proportional to x.
Two particles A and B possessing charges of +2.00 × 10−6 C and of −4.00 × 10−6 C, respectively, are held fixed at a separation of 20.0 cm. Locate the points (s) on the line AB, where (a) the electric field is zero (b) the electric potential is zero.
A water particle of mass 10.0 mg and with a charge of 1.50 × 10−6 C stays suspended in a room. What is the magnitude of electric field in the room? What is its direction ?
Two equal charges, 2.0 × 10−7 C each, are held fixed at a separation of 20 cm. A third charge of equal magnitude is placed midway between the two charges. It is now moved to a point 20 cm from both the charges. How much work is done by the electric field during the process?
Solve numerical example.
Three equal charges of 10×10-8 C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm, and 25cm respectively. Find the force exerted on the charge located at the 90° angle.
A force F acts between sodium and chlorine ions of salt (sodium chloride) when put 1 cm apart in air. The permittivity of air and dielectric constant of water are `epsilon_0` and K respectively. When a piece of salt is put in water, electrical force acting between sodium and chlorine ions 1 cm apart is ____________.
The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become ____________.
The electric force acting between two point charges kept at a certain distance in vacuum is 16 N. If the same two charges are kept at the same distance in a medium of dielectric constant 8, the electric force acting between them is ____________ N.
Polarised dielectric is equivalent to ______.
Two positive charges ______.
Two charges of equal magnitudes kept at a distance r exert a force F on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is ______.
There is another useful system of units, besides the SI/mks A system, called the cgs (centimeter-gram-second) system. In this system Coloumb’s law is given by
F = `(Qq)/r^2 hatr`
where the distance r is measured in cm (= 10–2 m), F in dynes (= 10–5 N) and the charges in electrostatic units (es units), where 1 es unit of charge = `1/([3]) xx 10^-9 C`
The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by c = 2.99792458 × 108 m/s. An approximate value of c then is c = [3] × 108 m/s.
(i) Show that the coloumb law in cgs units yields
1 esu of charge = 1 (dyne)1/2 cm.
Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L.
(ii) Write 1 esu of charge = x C, where x is a dimensionless number. Show that this gives
`1/(4pi ∈_0) = 10^-9/x^2 (N*m^2)/C^2`
With `x = 1/([3]) xx 10^-9`, we have `1/(4pi ∈_0) = [3]^2 xx 10^9 (Nm^2)/C^2`
or, `1/(4pi ∈_0) = (2.99792458)^2 xx 10^9 (Nm^2)/C^2` (exactly).
