
The plot $E$ Vs $r$ can be properly represented by
(A)
(B)
(C)
(D) None of these
Answer
545.7k+ views
Hint : First of all we should know about the Electric field, hence Electric field is the region around a charged particle or object within which its influence or a force would be experienced or exerted on other charged particles or objects.
Electric field which is produced by the stationary charges, and the magnetic field is produced by the moving charges (currents); these two are often described as the sources of the field.
Complete step by step solution:
The electrostatic force per unit charge is termed as the Electric Field. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward From a positive charge the electric field is radially outward and radially in toward a negative point charge.
The electrostatic force exerted by a point charge on a test charge at a r distance mainly depends upon the charge on both the charges, as well as the distance between the two.
As we know, for a shell of a radius and Q charge, the electric field is given by-
For a point inside the shell is when $ r < a $ , the field E=0
For point on or outside the shell is when $ r > R $ the field $ E = \dfrac{{kq}}{{{r^2}}} $
Thus, the graph (A) is correct.
Note:
Electric field, an electric property associated with each point in space when charge is present in any form. Electric field’s direction and magnitude are expressed by the value E, which is called the electric field strength or electric field intensity or simply the electric field.
Electric field which is produced by the stationary charges, and the magnetic field is produced by the moving charges (currents); these two are often described as the sources of the field.
Complete step by step solution:
The electrostatic force per unit charge is termed as the Electric Field. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward From a positive charge the electric field is radially outward and radially in toward a negative point charge.
The electrostatic force exerted by a point charge on a test charge at a r distance mainly depends upon the charge on both the charges, as well as the distance between the two.
As we know, for a shell of a radius and Q charge, the electric field is given by-
For a point inside the shell is when $ r < a $ , the field E=0
For point on or outside the shell is when $ r > R $ the field $ E = \dfrac{{kq}}{{{r^2}}} $
Thus, the graph (A) is correct.
Note:
Electric field, an electric property associated with each point in space when charge is present in any form. Electric field’s direction and magnitude are expressed by the value E, which is called the electric field strength or electric field intensity or simply the electric field.
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