Advertisements
Advertisements
प्रश्न
Consider the situation of the previous problem. A charge of −2.0 × 10−4 C is moved from point A to point B. Find the change in electrical potential energy UB − UA for the cases (a), (b) and (c).
Advertisements
उत्तर
Given:
Magnitude of charge, q = −2.0 × 10−4 C
(a) The electric field is along the x direction.
Thus, potential difference between (0, 0) and (4, 2),
dV = −E.dx = −20 × 4 = −80 V
Potential energy (UB − UA) between the points = dV × q
⇒ UB − UA = (−80) × (−2.0 × 10−4)
⇒ UB − UA = 160 × 10−4 = 0.016 J
(b) A = (4 m, 2m), B = (6 m, 5 m)
dV = −E.dx = − 20 × 2 = −40 V
Potential energy (UB − UA) between the points = dV × q
⇒ UB − UA = (−40) × (−2 × 10−4)
⇒ UB − UA = 80 × 10−4 = 0.008 J
(c) A = (0, 0) B = (6m, 5m)
dV = −E.dx = −20 × 6 = −120 V
Potential energy (UB − UA) between the points A and B = dV × q
⇒ UB − UA = (−120) × (−2 × 10−4)
⇒ UB − UA = 240 × 10−4 = 0.024 J
APPEARS IN
संबंधित प्रश्न
Show that if we connect the smaller and the outer sphere by a wire, the charge q on the former will always flow to the latter, independent of how large the charge Q is.
Consider a system of n charges q1, q2, ... qn with position vectors `vecr_1,vecr_2,vecr_3,...... vecr_n`relative to some origin 'O'. Deduce the expression for the net electric field`vec E` at a point P with position vector `vecr_p,`due to this system of charges.
A hollow cylindrical box of length 1 m and area of cross-section 25 cm2 is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by `vecE = 50xhati` where E is NC−1 and x is in metres. Find
(i) Net flux through the cylinder.
(ii) Charge enclosed by the cylinder.
The charge on a proton is +1.6 × 10−19 C and that on an electron is −1.6 × 10−19 C. Does it mean that the electron has 3.2 × 10−19 C less charge than the proton?
Can a gravitational field be added vectorially to an electric field to get a total field?
Why does a phonograph record attract dust particles just after it is cleaned?
Electric potential decreases uniformly from 120 V to 80 V, as one moves on the x-axis from x = −1 cm to x = +1 cm. The electric field at the origin
(a) must be equal to 20 Vcm−1
(b) may be equal to 20 Vcm−1
(c) may be greater than 20 Vcm−1
(d) may be less than 20 Vcm−1
Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum.
A particle of mass 1 g and charge 2.5 × 10−4 C is released from rest in an electric field of 1.2 × 10 4 N C−1. How much is the work done by the electric force on the particle during this period?
An electric field \[\vec{E} = ( \vec{i} 20 + \vec{j} 30) {NC}^{- 1}\] exists in space. If the potential at the origin is taken to be zero, find the potential at (2 m, 2 m).
An electric field \[\vec{E} = \vec{i}\] Ax exists in space, where A = 10 V m−2. Take the potential at (10 m, 20 m) to be zero. Find the potential at the origin.
The electric potential existing in space is \[\hspace{0.167em} V(x, y, z) = A(xy + yz + zx) .\] (a) Write the dimensional formula of A. (b) Find the expression for the electric field. (c) If A is 10 SI units, find the magnitude of the electric field at (1 m, 1 m, 1 m).
Find the magnitude of the electric field at the point P in the configuration shown in the figure for d >> a.
Five charges, q each are placed at the corners of a regular pentagon of side ‘a’ (Figure).
(a) (i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the corners (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by –q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its corners?
