Two tiny particles having charges +20.0E-6 C and -8.00E-6 C ...

Question

Two tiny particles having charges +20.0E-6 C and -8.00E-6 C are separated by a distance of 20.0 cm. What are the magnitude and direction of electric field midway between these two charges?

1.  A. 25.2E6 N/C directed towards the negative charge
2.  B. 25.2E6 N/C directed towards the positive charge
3.  C. 25.2E5 N/C directed towards the negative charge
4.  D. 25.2E5 N/C directed towards the positive charge
5.  E. 25.2E4 N/C directed towards the negative charge

Answer 1

Since this is a multiple choice, we can rule out some answers in our head. The distance between a charge and the mid point is 10 cm (0.1 m). Since they are the same distance apart, the field from the larger charge will be stronger. Therefore, we know the electric field will be pointing to the positive 20E-6 C particle. That leaves only B and D as answers. Now we just need to determine the magnitude of the electric field. The magnitude due to one charge is E = kQ/d² where k is a constant (8.987E9 Nm²/C²), Q is the charge, and d is the distance to the test point. The key point to know, is that two electric fields add together when combined so we have the following equation: E_positve + E_negative = E_total. Using the formula for E we now have k*(20E-6C/(0.1m)²) + k*(-8E-6C/(0.1m)²) = k*(.002C/m² - 0.0008C/m²) = k*(0.0012C/m²)

~ 9E9 Nm²/C² * (0.0012C/m²) = 10.8E6 N/C.

Here we see that NONE of the answers are correct! We can see the mistake that was made though. If we used a POSITVE 8e-6 C as a charge we would have 9E9 Nm²/C² * (0.0028C/m²) = 25.2E6 N/C, which matches answer B.

It's very important to know that the sign of the electric field matters and cannot be ignored.