<P> This is the electric field at point x 0 (\ displaystyle (\ boldsymbol (x_ (0)))) due to the point charge q 1 (\ displaystyle q_ (1)); it is a vector equal to the Coulomb force per unit charge that a positive point charge would experience at the position x 0 (\ displaystyle (\ boldsymbol (x_ (0)))). Since this formula gives the electric field magnitude and direction at any point x 0 (\ displaystyle (\ boldsymbol (x_ (0)))) in space (except at the location of the charge itself, x 1 (\ displaystyle (\ boldsymbol (x_ (1)))), where it becomes infinite) it defines a vector field . From the above formula it can be seen that the electric field due to a point charge is everywhere directed away from the charge if it is positive, and toward the charge if it is negative, and its magnitude decreases with the inverse square of the distance from the charge . </P> <P> If there are multiple charges, the resultant Coulomb force on a charge can be found by summing the vectors of the forces due to each charge . This shows the electric field obeys the superposition principle: the total electric field at a point due to a collection of charges is just equal to the vector sum of the electric fields at that point due to the individual charges . </P> <Dl> <Dd> E (x) = E 1 (x) + E 2 (x) + E 3 (x) + ⋯ = 1 4 π ε 0 q 1 (x 1 − x) 2 r ^ 1 + 1 4 π ε 0 q 2 (x 2 − x) 2 r ^ 2 + 1 4 π ε 0 q 3 (x 3 − x) 2 r ^ 3 + ⋯ (\ displaystyle (\ boldsymbol (E)) ((\ boldsymbol (x))) = (\ boldsymbol (E_ (1))) ((\ boldsymbol (x))) + (\ boldsymbol (E_ (2))) ((\ boldsymbol (x))) + (\ boldsymbol (E_ (3))) ((\ boldsymbol (x))) + \ cdots = (1 \ over 4 \ pi \ varepsilon _ (0)) (q_ (1) \ over ((\ boldsymbol (x_ (1) - x))) ^ (2)) (\ boldsymbol ((\ hat (r)) _ (\ text (1)))) + (1 \ over 4 \ pi \ varepsilon _ (0)) (q_ (2) \ over ((\ boldsymbol (x_ (2) - x))) ^ (2)) (\ boldsymbol ((\ hat (r)) _ (\ text (2)))) + (1 \ over 4 \ pi \ varepsilon _ (0)) (q_ (3) \ over ((\ boldsymbol (x_ (3) - x))) ^ (2)) (\ boldsymbol ((\ hat (r)) _ (\ text (3)))) + \ cdots) </Dd> <Dd> E (x) = 1 4 π ε 0 ∑ k = 1 N q k (x k − x) 2 r ^ k (\ displaystyle (\ boldsymbol (E)) ((\ boldsymbol (x))) = (1 \ over 4 \ pi \ varepsilon _ (0)) \ sum _ (k = 1) ^ (N) (q_ (k) \ over ((\ boldsymbol (x_ (k) - x))) ^ (2)) (\ boldsymbol ((\ hat (r)) _ (\ text (k))))) </Dd> <Dd> where r ^ k (\ displaystyle (\ boldsymbol ((\ hat (r)) _ (\ text (k))))) is the unit vector in the direction from point x k (\ displaystyle (\ boldsymbol (x_ (k)))) to point x (\ displaystyle (\ boldsymbol (x))). </Dd> </Dl> <Dd> E (x) = E 1 (x) + E 2 (x) + E 3 (x) + ⋯ = 1 4 π ε 0 q 1 (x 1 − x) 2 r ^ 1 + 1 4 π ε 0 q 2 (x 2 − x) 2 r ^ 2 + 1 4 π ε 0 q 3 (x 3 − x) 2 r ^ 3 + ⋯ (\ displaystyle (\ boldsymbol (E)) ((\ boldsymbol (x))) = (\ boldsymbol (E_ (1))) ((\ boldsymbol (x))) + (\ boldsymbol (E_ (2))) ((\ boldsymbol (x))) + (\ boldsymbol (E_ (3))) ((\ boldsymbol (x))) + \ cdots = (1 \ over 4 \ pi \ varepsilon _ (0)) (q_ (1) \ over ((\ boldsymbol (x_ (1) - x))) ^ (2)) (\ boldsymbol ((\ hat (r)) _ (\ text (1)))) + (1 \ over 4 \ pi \ varepsilon _ (0)) (q_ (2) \ over ((\ boldsymbol (x_ (2) - x))) ^ (2)) (\ boldsymbol ((\ hat (r)) _ (\ text (2)))) + (1 \ over 4 \ pi \ varepsilon _ (0)) (q_ (3) \ over ((\ boldsymbol (x_ (3) - x))) ^ (2)) (\ boldsymbol ((\ hat (r)) _ (\ text (3)))) + \ cdots) </Dd>

What is the strength of an electric field generated by charge at a distance of 1 m