Question

At what distance from a point charge 5x10-8C, the
intensity of electric field be 450 V/m? (UP 2012)
Ans. Im​

Answer 1

Explanation:

q=5*10^-8C

E=450V/mor N/C

E=f/q

E=kq/r^2

E=9*10^9*5*10^-8/r^2

r^2=1

r=1

Answer 2

The distance from the point charge that will create an electric field of intensity 450 V/m will be 1 metre.

Given,

Magnitude of the point charge=$5X10^{-8} C$

Intensity of the electric field=450 V/m.

To find,

the distance from the point charge that will create an electric field of intensity 450 V/m.

Solution:

• Electric field strength at a point is defined as the electric force acting on a positive charge placed at that point divided by the test charge.
• $E=\frac{F}{q}$.
• Electric field, E due to a charge, q at a distance. r from it is given as:
• $E=\frac{kq}{r^{2} }$.
• $k=9X10^{9} N m^{2} /C^{2}$.
• Electric field is always away from a positive charge and towards a negative charge.
• Its unit is N/C or Volt/m.

The distance from the point charge that will create an electric field of intensity 450 V/m will be:

$E=\frac{kq}{r^{2} }\\450=\frac{9X10^{9} X5X10^{-8} }{r^{2} } \\450=\frac{45X10 }{r^{2} }\\450=\frac{450 }{r^{2} }\\r^{2} =1m\\r=1 m.$

Hence, the distance from the charge is 1 m.

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