Question

A negative charge of  2.0×10−8 C experiences a force of 0.060 N to the right in
an electric field.  What is the magnitude and direction of the field?
F1/q = F2/q

Answer 1

Explanation:

F = qE

0.06   = 2 x 10 ⁻⁸ xE

E = 3 x10⁶

E field provide the the force on the particle .Therefore direction of field is along the direction of force -Right

                                      BRAINLIEST                            

Answer 2

Answer:

3,000,000 N/C or 3 · $10^{6}$ N/C, to the left

Explanation:

To get the magnitude of the Electric Field, use the following formula:

$E = \frac{F}{q'}$

F = 0.060 N

q' = 2.0 · $10^{-8}$ C

$E = \frac{F}{q'} =\frac{0.060 N}{2.0 * 10^-8 C}= 3,000,000 N/C$

the scientific number of 3,000,000 is 3 · $10^{6}$

The direction of the field depends on the charge.

"Negative charges attract or radiate toward themselves."

"Positive charges radiate outwards or against themselves."

Since it has a negative charge, it attracts. If the force was experienced to the right of the negative charge, the electric field is going to the left.