An electric field exerts a force of 3.00 E-4 N on a positive test charge of 7.20 E-4 C. The magnitude of the field at the location of the charge is?
The field,. designated by E, satisfies the formula
F = Q*E
Therefore E = F/Q
The units will be Newtons per Coulomb, which is the same as Volts per meter
To find the magnitude of the electric field, we can use the formula:
Electric field (E) = Force (F) / Test charge (q)
Given:
Force (F) = 3.00 E-4 N
Test charge (q) = 7.20 E-4 C
Substituting these values into the formula, we get:
Electric field (E) = 3.00 E-4 N / 7.20 E-4 C
Now, simplify the expression:
Electric field (E) = 0.4167 N/C
Therefore, the magnitude of the electric field at the location of the charge is approximately 0.4167 N/C.
To find the magnitude of the electric field at the location of the charge, we can use the formula:
Electric Field (E) = Force (F) / Test Charge (q)
Given in the question, the force (F) exerted on the test charge is 3.00 E-4 N, and the test charge (q) is 7.20 E-4 C.
Plugging the values into the formula:
E = 3.00 E-4 N / 7.20 E-4 C
Now we can simplify the equation:
E = (3.00 / 7.20) * (10^-4 / 10^-4) N/C
E = 0.4167 * 1 N/C
Finally, we get:
E = 0.4167 N/C
Therefore, the magnitude of the electric field at the location of the charge is 0.4167 N/C.