Respuesta :
Answer:
work done by electric field is 0.06 J
Explanation:
Given data:
Two point charge is [tex]+ 1\mu C and -1 \mu C[/tex]
0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)
-1 charge positioned is (0 cm , -1 cm, 0.00 cm)
E = 3.0\times 10^6 N/C
From above information, the distance between given two charges d = 2 cm
then d = 0.02m
work needed is W = q E d
[tex]W = 1.0 \times 10^{-6} \times 3.0 \times 10^6 \times 0.02[/tex]
W = 0.06 J
Therefore work done by electric field is 0.06 J
The calculated work value is "0.06 J".
Work calculation:
For positive charge:
[tex]\to q = 1.0 \mu C[/tex]
located at [tex]\ ( x , y , z ) = ( 0 , 1cm , 0 )[/tex]
For negative charge:
[tex]\to q = - 1.0 \mu C[/tex]
located at [tex]( x , y , z ) = ( 0 , -1 cm , 0 )[/tex]
[tex]\to E = ( 3 \times 10^6 \ \frac{N}{C} ) \ i[/tex]
using given data the distance between two charges [tex]d = 2\ cm[/tex]
then [tex]d = 0.02\ m[/tex]
Using a formula for calculating the Work:
[tex]\to W = q E d[/tex]
[tex]= 1.0 \times 10^{-6} \times 3 \times 10^6 \times 0.02\\\\= 1.0 \times 3 \times 0.02\\\\=0.06 \ J[/tex]
Therefore, the final answer is "0.06 J".
Find out more about the electric dipole here:
brainly.com/question/14552607
