A+5.0-μc point charge is placed at the 0 cm mark of a meter stick and a -4.0-μc charge is placed at the 50 cm mark. what is the net electric field at the 30 cm mark? (k = 1/4πε0 = 8.99 × 109 n • m2/c2) a +5.0-μc point charge is placed at the 0 cm mark of a meter stick and a -4.0-μc charge is placed at the 50 cm mark. what is the net electric field at the 30 cm mark? (k = 1/4πε0 = 8.99 × 109 n • m2/c2) 4.0 × 105 n/c 1.4 × 106 n/c 9.0 × 105 n/c 5.0 × 105 n/c

1.4 *10^6 N/C

Explanation:

The electric field caused by a charge at a certain point is given by the equation:

where k is the Coulomb constant equal to 8.99 *10^9 Nm^2/C^2, q the charge of the particle in coulombs, r is the distance from the point to the charge in meters.

is the unitary vector that goes from the charge to the point. This vector will give us the direction of the Electric Field vector.

The unitary vector of the +5.0-μC charge will go to the right (+i), as the point is to the right of the charge. Then, the electric field caused by the charge will be:

The unitary vector of the -4.0-μC charge will go to the left (-i), as the point is to the left of the charge. Then, the electric field caused by the charge will be:

The electric field at the 30 cm mark will be the addition of both electric field:

the answer is false

explanation:

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