Consider a dipole with charges q1 = +q
and q2 = –q placed in a uniform electric field E, as shown in Fig.
In a uniform electric field, the dipole experiences no net force; but
experiences a
Torque =p×E
(which will tend to rotate it
(unless p is parallel or anti-parallel to E). Suppose an external torque ext is applied in such a manner that it just neutralizes
this torque and rotates it in the plane of paper from angle θ0 to
angle θ1 at an infinitesimal angular speed and without angular
acceleration. The amount of work done by the external torque will be given by
This work is stored as the potential energy of the
system. A natural choice is to take θ0 = π / 2. (Αn explanation for
it is provided towards the end of discussion.) We can then write,
Here, r1 and r2 denote the
position vectors of +q and –q. Now, the potential difference between positions
r1 and r2 equals the work done in bringing a unit
positive charge against field from r2 to r1. The
displacement parallel to the force is 2a cosθ. Thus, [V (r1)–V (r2)]
= –E × 2a cosθ. We thus obtain,
_________________________________________________________________________________
Some of these questions which may be asked in your Board Examination 2012-2013
Q1: what are the characteristic of charges acquired by the objects on rubbing against each other?
Q2: Who suggested first that there are two kinds of charges?
Q3: How can you show that there are two types of charges?
Q4: An ebonite rod is rubbed with the fur or wool. What type of charges do they acquire?
Q5: Is mass of body affected on charging?
Q6: What is the polarity of charge?
Answer these questions in comment box and help your friends
Q1: what are the characteristic of charges acquired by the objects on rubbing against each other?
Q2: Who suggested first that there are two kinds of charges?
Q3: How can you show that there are two types of charges?
Q4: An ebonite rod is rubbed with the fur or wool. What type of charges do they acquire?
Q5: Is mass of body affected on charging?
Q6: What is the polarity of charge?
Answer these questions in comment box and help your friends
No comments:
Post a Comment