Question

If the electric field is 100N/C at a distance of 50 cm from a point charge , what is the value of ?.

Answer 1

Answer:

$e \: = \: \frac{kq}{{r}^{2} }$

then q = e r² / k

then q = 100 * (50 ×10^-2)² / 9×10^9 = 2.777777778×10^-9 C

Answer 2

$2.8\times 10^{-9}C$ is value of q

What is electric field?

An electric property associated with each point in space when charge is present in any form.

The magnitude and direction of the electric field are expressed by the value of E, called electric field strength .

By definition,

The electric field of a point charge is given as follows:

$E = k\dfrac{q}{r^2}$

​Where q is the charge,

r = 50cm = 0.5mr=50cm=0.5m is the distance from the charge,

k =$9\times 10^9N\cdot m^2/C^2$

C is the Coulomb's constant.

Substituting E = 100N/C and expressing q,

obtain:

$q = \dfrac{Er^2}{k} = \dfrac{100\cdot 0.5^2}{9\times 10^9} \approx 2.8\times 10^{-9}C$

Therefore,

$2.8\times 10^{-9}C$ is the value of q.

The complete question is given below:

If the electric field is 100N/C at a distance of 50 cm from a point charge , what is the value of q?

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