Advertisements
Advertisements
प्रश्न
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?
उत्तर
Let the third charge be moved from point A to point B.
Magnitude of the charges, q = 2.0 × 10−7 C
Distance of both the charges from point A, r = 10 cm =0.1 m
Distance of both the charges from point B, r' = 20 cm =0.2 m
Potential at A,
\[V_A = 2 \times \frac{1}{4\pi \epsilon_0}\frac{q}{r}\]
Potential at B,
\[V_B = 2 \times \frac{1}{4\pi \epsilon_0}\frac{q}{r'}\]
Work done,
\[\Rightarrow W = 2 \times {10}^{- 7} \left[ \frac{2 \times 2 \times {10}^{- 7}}{4 \pi\epsilon_0}\left( \frac{1}{0 . 1} - \frac{1}{0 . 2} \right) \right]\]
\[ \Rightarrow W = 3 . 6 \times {10}^{- 3}\] J
APPEARS IN
संबंधित प्रश्न
Four point charges qA = 2 μC, qB = −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
- Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation.
- What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
A charge of 1.0 C is placed at the top of your college building and another equal charge at the top of your house. Take the separation between the two charges to be 2.0 km. Find the force exerted by the charges on each other. How many times your weight is this force?
At what separation should two equal charges, 1.0 C each, be placed, so that the force between them equals the weight of a 50 kg person?
NaCl molecule is bound due to the electric force between the sodium and the chlorine ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be 2.75 × 10−8 cm, find the force of attraction between them. State the assumptions (if any) that you have made.
Ten positively-charged particles are kept fixed on the x-axis at points x = 10 cm, 20 cm, 30 cm, ...., 100 cm. the first particle has a charge 1.0 × 10−8 C, the second 8 × 10−8 C, the third 27 × 10−8 C and so on. The tenth particle has a charge 1000 × 10−8 C. Find the magnitude of the electric force acting on a 1 C charge placed at the origin.
Answer the following question.
What is relative permittivity?
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.
The ratio of the forces between two small spheres with constant charge (a) in air (b) in a medium of dielectric constant K is ______.
For charges q1 and q2 separated by a distance R the magnitude of the electrostatic force is given by ______.
A spring of spring constant 5 × 103 N/m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is:
Two charge – 10c and + 10 c are placed 10 cm apart. Potential at centre of the line joining the two charge is:-
Electric charge of any system is ______.
Two charges q and – 3q are placed fixed on x-axis separated by distance ‘d’. Where should a third charge 2q be placed such that it will not experience any force?
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).
According to Coulomb's law, which is the correct relation for the following figure?
Which of the following statements about nuclear forces is not true?
The ratio of the forces between two charges placed at a certain distance apart in the air and by the same distance apart in a medium of dielectric constant K is ______.
