As charge of particle 2 mc is at the position (1,1,1) m the electrostatic force on another chargedaeticle at 1 mc at (0,0,0) .is

500N

6000N

400N

200N

Open in App

Solution

The distance between the two points are

$square root of open parentheses 1 minus 0 close parentheses squared plus open parentheses 1 minus 0 close parentheses squared plus open parentheses 1 minus 0 close parentheses squared end root equals square root of 3 equals d D squared equals 3 a n d space w e space k n o w space F equals fraction numerator k q subscript 1 q subscript 2 over denominator D squared end fraction g i v e n space q subscript 1 equals 2 x 10 to the power of negative 3 end exponent q subscript 2 equals 1 x 10 to the power of negative 3 end exponent F equals fraction numerator 9 cross times 10 to the power of 9 cross times 2 cross times 10 to the power of negative 3 end exponent cross times 10 to the power of negative 3 end exponent over denominator 3 end fraction F equals fraction numerator 6000 N over denominator blank end fraction$

Suggest Corrections

Similar questions

A charge particle $q_{1}$ is at position (2,-1,3).The electrostatic force on another charged particles $q_{2}$ at

(2, - 1, 3)

q_{2}

1 C

P(1, 2, 3)

2 C

Q(2, 3, 4)

(K = \frac{1}{4 π ϵ_{0}})

O

+ Q

+ Q

8 N .

+ Q

Join BYJU'S Learning Program