
Two infinitely long line charges each of linear charge density $ \lambda $ are placed at an angle $ \theta $ as shown in the figure. Find out electric field intensity at a point $ P $ , which is at a distance $ x $ from point $ O $ along the angle bisector of line charges.
Answer
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Hint: The basic physical property of matter that causes it to experience a force when kept in an electric or magnetic field is electrical charge. An electric charge is correlated with an electric field and a magnetic field is generated by the moving electric charge. The electromagnetic field is recognized as a combination of electrical and magnetic fields.
Formula Used: We will use the following formula:
$ E = \dfrac{{2k\lambda }}{r} $
Where
$ E $ is the electrical charge
$ \lambda $ is the linear charge density
$ r $ is perpendicular distance of the point from the line charge
$ k $ is the Coulomb’s constant.
Complete step by step solution:
Let us consider an infinitely long line charge whose linear charge density is $ \lambda $
So, the magnitude of the electric field at any point at a distance of $ r $ units will be
$ E = \dfrac{{2k\lambda }}{r} $
And the direction of this electric field will be away from the wire
According to the question, it is given that $ OP = x $
Let us assume that the length $ AP = PB = r $
Now according to the figure,
$ x\sin \dfrac{\theta }{2} = r $
Now we know that
$ E = \dfrac{{2k\lambda }}{r} $
Now, we will replace the value of $ r $ from the above expression. So, we will get
$ E = \dfrac{{2k\lambda }}{{x\sin \dfrac{\theta }{2}}} $
So, the net electrical charge will be
$ {E_{net}} = 2E\sin \dfrac{\theta }{2} $
$ = 2\left( {\dfrac{{2k\lambda }}{{x\sin \dfrac{\theta }{2}}}} \right) \times \sin \dfrac{\theta }{2} $
Therefore, the magnitude of the net electrical charge will be
$ {E_{net}} = \dfrac{{4k\lambda }}{x} $
And the direction of this net charge will be along $ OP $ .
Note:
An electrical charge is a scalar quantity. In addition to having a magnitude and direction, the laws of vector addition such as triangle law of vector addition and parallelogram law of vector addition should also obey a quantity to be called a vector, only then is the quantity said to be a quantity of vector. When two currents meet at a junction, in the case of an electric current, the resulting current will be an algebraic sum and not the sum of the vector. Therefore, a scalar quantity is an electric current, although it has magnitude and direction.
Formula Used: We will use the following formula:
$ E = \dfrac{{2k\lambda }}{r} $
Where
$ E $ is the electrical charge
$ \lambda $ is the linear charge density
$ r $ is perpendicular distance of the point from the line charge
$ k $ is the Coulomb’s constant.
Complete step by step solution:
Let us consider an infinitely long line charge whose linear charge density is $ \lambda $
So, the magnitude of the electric field at any point at a distance of $ r $ units will be
$ E = \dfrac{{2k\lambda }}{r} $
And the direction of this electric field will be away from the wire
According to the question, it is given that $ OP = x $
Let us assume that the length $ AP = PB = r $
Now according to the figure,
$ x\sin \dfrac{\theta }{2} = r $
Now we know that
$ E = \dfrac{{2k\lambda }}{r} $
Now, we will replace the value of $ r $ from the above expression. So, we will get
$ E = \dfrac{{2k\lambda }}{{x\sin \dfrac{\theta }{2}}} $
So, the net electrical charge will be
$ {E_{net}} = 2E\sin \dfrac{\theta }{2} $
$ = 2\left( {\dfrac{{2k\lambda }}{{x\sin \dfrac{\theta }{2}}}} \right) \times \sin \dfrac{\theta }{2} $
Therefore, the magnitude of the net electrical charge will be
$ {E_{net}} = \dfrac{{4k\lambda }}{x} $
And the direction of this net charge will be along $ OP $ .
Note:
An electrical charge is a scalar quantity. In addition to having a magnitude and direction, the laws of vector addition such as triangle law of vector addition and parallelogram law of vector addition should also obey a quantity to be called a vector, only then is the quantity said to be a quantity of vector. When two currents meet at a junction, in the case of an electric current, the resulting current will be an algebraic sum and not the sum of the vector. Therefore, a scalar quantity is an electric current, although it has magnitude and direction.
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