Question

an electric field strength created by charge q is measured to be 40 n/c at a distance of 0.2 m from the center of the charge. What is the new field strength when the distance from the center of Q is changed to 0.4m away with twice the charge Q​

Answer 1

Explanation:

• E=40n/c
• q=q
• r=0.2
• E=$\frac{kq}{{r}^{2} }$

now,

• q'=2q
• r'=0.4=2r
• E'=?

E'=kq'/(r'^2)

=k2q/(2r)^2

=k2q/4(r^2)

=kq/2(r^2)

=E/2

=40/2

=20n/c

hope it helps you...thank you..

Answer 2

At a distance of 0.4 m, the new electric field strength produced by the charge Q = 2q is 44.5 N/C.

Coulomb's law gives the following values for the electric field strength produced by a point charge q at a distance r from its centre:

$E = k\times q/r^2$

where

k is the Coulomb constant, which is equal to nearly

$9 \times 10^9 N-m^2/C^2.$

The electric field strength produced by the charge q at a distance of 0.2 m in this case is indicated to be 40 N/C.

With the use of this knowledge, we can determine what q is worth as follows:

$40 N/C = k\times q/(0.2 m)^2$

$q = (40 N/C)\times (0.2 m)^2/k$

$q = (40 N/C)\times\frac{0.04 m^2}{9 \times 10^9 Nm^2/C^2}$

$q = 1.78 \times 10^{-7} C$

The new electric field strength produced by a charge Q = 2q at a distance of 0.4 m may now be calculated using Coulomb's law:

$E' = k\times (2q)/(0.4 m)^2$

$E' = (2kq)/(0.4 m)^2$

$E' = (2\times(9 \times 10^9 Nm^2/C^2)\times(1.78 \times10^{-7} C))/(0.4 m)^2$

$E' = 44.5\: N/C$

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