हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

The Electric Potential Existing in Space is V ( X , Y , Z ) = a ( X Y + Y Z + Z X ) . (A) Write the Dimensional Formula of A.(B) Find the Expression for the Electric Field. - Physics

Advertisements
Advertisements

प्रश्न

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).

संख्यात्मक
Advertisements

उत्तर

Given:
Electric potential, 

\[V(x, y, z) = A(xy + yz + zx)\]

\[A = \frac{\text{ volt }}{m^2}\] 

\[ \Rightarrow \left[ A \right] = \frac{\left[ {ML}^2 I^{- 1} T^{- 3} \right]}{\left[ L^2 \right]}\] 

\[ \Rightarrow A = [ {MT}^{- 3}  I^{- 1} ]\]

(b) Let be the electric field.

\[dV =  -  \vec{E}  .  \vec{dr} \] 

\[ \Rightarrow A(y + z)dx + A(z + x)dy + A(x + y)dz =  - E(dx \hat{i}  + dy \hat{j}  + dz\hat{ k } )\] 

\[ \Rightarrow [A(y + z) \hat{i }  + A(z + x)\hat{ j }  + A(x + y) \hat{ k } ]  [dx\hat{ i}  + dy \hat{j }  + dz \hat{k } ] =  - E\left[ dx \hat{ i }+ dy\hat{ j } + dz \hat{ k } \right]\]

Equating now, we get

\[\vec{E}  =  - A(y + z) \hat{ i }  - A(z + x) \hat{ j }  - A(x + y) \hat{ k }\]

(c) Given: A = 10 V/m2

\[r = (1  m,   1  m,   1  m)\] 

\[ \vec{E}  =  - 10  (2) \hat{ i }  - 10  (2) \hat{ j } - 10  (2) \hat{ k } \] 

\[       =  - 20 \hat{ i } - 20  \hat{ j }  - 20 \hat{ k }\]

Magnitude of electric field,

\[\left| E \right| = \sqrt{{20}^2 + {20}^2 + {20}^2}\] 

\[ = \sqrt{1200} = 34 . 64 = 35\] N/C 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Electric Field and Potential - Exercises [पृष्ठ १२३]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 7 Electric Field and Potential
Exercises | Q 60 | पृष्ठ १२३

संबंधित प्रश्न

A hollow cylindrical box of length 0.5 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 = 20 xhati`  where E is NC­−1 and x is in metres. Find

(i) Net flux through the cylinder.

(ii) Charge enclosed by the cylinder.


Can a gravitational field be added vectorially to an electric field to get a total field?


If a body is charged by rubbing it, its weight


A point charge q is rotated along a circle in an electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is 


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 


A wire is bent in the form of a regular hexagon and a total charge q is distributed uniformly on it. What is the electric field at the centre? You may answer this part without making any numerical calculations. 


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 long will it take for the particle to travel a distance of 40 cm?


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). 


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.


Find the magnitude of the electric field at the point P in the configuration shown in the figure for d >> a.


Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire of mass 6.4 g (take the atomic weight of copper to be 64 g mol−1). 


The unit of electric field is not equivalent to ______.

A charged particle is free to move in an electric field. It will travel ______.

Consider a region inside which, there are various types of charges but the total charge is zero. At points outside the region ______. 


When 1014 electrons are removed from a neutral metal sphere, the charge on the sphere becomes ______.


The electric field intensity produced by the radiations coming from 100 W bulb at 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is:


The Electric field at a point is ______.

  1. always continuous.
  2. continuous if there is no charge at that point.
  3. discontinuous only if there is a negative charge at that point.
  4. discontinuous if there is a charge at that point.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×