Understanding Electric Fields between Charges

[roy5995]

Originally posted by marshall4
Charge A is +2.5*10^-5C Charge B is -3.7*10^-7C.
A is 25cm away from B. Point C is 10cm away from point B. What is the electric field at C? All points are in a line A--B-C.

Would this work?

Find the electric fields a pont A & B. Then add the two fields ? Or would you subtract the two fields ?

Use the formula {E} = k\frac{Q}{r^2}


Is the field on point A & B going left or right?

By Reasoning does the electric field at A have to be bigger that B, since A is further away, Or should B be bigger that A , or is there no way of telling, just by looking at the question

 

[Adrian Baker]

Originally posted by roy5995
Would this work?

Find the electric fields a pont A & B. Then add the two fields ? Or would you subtract the two fields ?

Use the formula {E} = k\frac{Q}{r^2}


Is the field on point A & B going left or right?

By Reasoning does the electric field at A have to be bigger that B, since A is further away, Or should B be bigger that A , or is there no way of telling, just by looking at the question

In reply to the first part of your question, electric field is a vector quantity. You need to think what is the direction the field due to each charge. Try drawing it.

As one is +ve and the other -ve, both would move a positive charge in the same direction, so you add the values.

You say "Find the electric fields a pont A & B.". Actually you should find the value of the electric fields AT C due to A and at C due to B.

 

[M98Ranger]

Thank you for your question. Firstly, to find the electric field at point C, we need to use the formula {E} = k\frac{Q}{r^2}, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), Q is the charge in Coulombs, and r is the distance in meters.

To find the electric field at point A, we will use the charge of +2.5*10^-5C and the distance of 25cm (or 0.25m). Plugging these values into the formula, we get {E}_A = (9x10^9)(2.5*10^-5)/(0.25)^2 = 0.9x10^6 N/C.

Similarly, to find the electric field at point B, we will use the charge of -3.7*10^-7C and the distance of 0.1m. Plugging these values into the formula, we get {E}_B = (9x10^9)(-3.7*10^-7)/(0.1)^2 = -3.33x10^5 N/C.

Now, to find the electric field at point C, we need to add the two fields at points A and B. This is because the electric fields at these points will be acting in the same direction, since they are both positive charges. So, the electric field at point C will be {E}_C = {E}_A + {E}_B = 0.9x10^6 N/C - 3.33x10^5 N/C = 5.67x10^5 N/C.

Using this method, we can determine that the electric field at point C is 5.67x10^5 N/C.

To answer your other questions, the electric field at point A will be going to the right, as the charge is positive and the electric field lines always point away from positive charges. Similarly, the electric field at point B will be going to the left, as the charge is negative and the electric field lines always point towards negative charges.

As for the size of the electric fields at points A and B, we cannot determine which one will be bigger just by looking at the question. We need to calculate the electric fields using the formula and then compare them to determine which one is