Question

calculate the electric field intensity over the charge at a point separated from source charge of +5 micro coulomb by a distance of 5 cm​

Answer 1

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Answer 2

Answer:

The electric field intensity over the charge at a point separated from a source charge of +5 micro coulomb by a distance of 5 cm​ is $9*10^{-5}N/C$

Explanation:

When two charges $q_{1} ,q_{2}$ are separated by a distance $r$ they will interact with each other via electrostatic force, which is given by the formula $F=K\frac{q_{1} q_{2} }{r^{2} }$

where $K=9*10^{9}$

the electric field is the electrostatic force per unit charge (that is $q_{2}=1C$)

It can be expressed as $E=K\frac{q_{1} }{r^{2} }$

From the question, it is given that the charge $q_{1}=5*10^{-6}C$

and the distance is $d=5cm=0.05m$

now by substituting the given values we get

electric field is $E=\frac{9*10^{9}*5*10^{-6} }{0.05^{2} }=9*10^{-5}N/C$