Question

Two charges Q =2 C and q=1 nC are 1 m apart. If the electric force beween them is 18 N what is the magnitude of the electric field due to Q at q. The format of the answer must be g

Answer 1

magnitude of the electric field = 1.8 ×
`10^(10)` N/C

Step-by-step explanation:

given data

charge Q =2 C

charge q = 1 nC = 1 ×
`10^(-9)` C

distance r = 1 m

electric force = 18 N

solution

we know that here electric field due to Q is

E = k ×
`(Q)/(r^2)` ..............1

here q distance is 1 m

now we apply here Coulomb’s law and get here electric force

F = k ×
`(Q*q)/(r^2)` ..........2

we know constant k = 8.988 ×
`10^(9)` Nm²/C²

so from above both equation we get

electric filed =
`(force)/(charge)` .................3

put here value

electric filed =
`(18)/(1*10^(-9))` = 1.8 ×
`10^(10)` N/C

Answer 2

`E=1.8* 10^(10)\ N.C^(-1)`

Step-by-step explanation:

Given:

• one charge,
  `Q=2\ C`

• another charge,
  `q=10^(-9)\ C`

• distance between the two charges,
  `r=1\ m`

• force between the charges,
  `F=18\ N`

We know from the Coulomb's law:

`F=(1)/(4\pi.\epsilon_0) * (Q.q)/(r^2)` ...........(1)

where:

`\epsilon_0=` permittivity of free space

Also we have electric field due to Q at q (which is at a distance of 1 m):

`E=(1)/(4\pi.\epsilon_0) * (Q)/(r^2)\ [N.C^(-1)]` ...........(2)

From (1) & (2)

`E=(F)/(q)`

`E=(18)/(10^(-9))`

`E=1.8* 10^(10)\ N.C^(-1)`