Recent content by paulimerci

The ratio of ##\frac{E_P}{E_R}## gives,

ok. The electric field at point R is $$ E_{R} = \frac{kq}{100}$$ The electric field at point P is $$ E_{p} = \frac {kq}{9}$$ $$ 9\times E_{P} = kq$$ $$ 9\times E_{P} = E_R \times 100$$ $$ E_R = \frac {9 E_P} {100}$$ Is it right?

Thank you!

I've no idea how to solve this problem. The sign of the charge is not mentioned, so I'm assuming the charge is "+". The charge exerts an outward electric field. Since two lengths of the right-angle triangle are given, I use the Pythagorean to find the hypotenuse, which is the distance between q...

I see where I did a mistake I took the wrong distance (between P and the midpoint between two charges). r=3m, Is ##E_{1y} + E_{2y} = 2598+2598 = 5196 N/C##. So the answer choice is B?

r is the distance between the charge and the reference point (P).

There are two identical spheres with the same charge that are the vertices of an equilateral triangle. ##+3 \mu C## will exert an outward electric field, which is drawn in the FBD below (see the attached pic), Since the horizontal force components (1x and 2x) are equal and opposite at point P...

The answer is close to 100 N. Thank you @TSny for your great help!

I'm sorry I used ## k= 9.1 \times 10^9 Nm^2/C^2## that's how I got the answer for ##F_{AP} = 91N ##

Yes, the distance between B and P is greater than the distance between A and P. I took ##k = 9\times 10^9 Nm^2/C^2## I did something wrong in calculating the distance. ##F_{BP} = 45.5 N ## and ##F_{AP} = 91N##. Thereby the magnitude of electric force at point P is 102N.

Ok. ##F_{AP} = 91N## ##F_{BP} = 91N## Using Pythagorean, the net magnitude of electric force at point p is 128N.

I think I understood what you said. So the magnitude of the electric force is 128N?

Thanks for the diagram. I can't visualize F_BP perpendicular to F_AP.

Isn’t the ##F_{BP} sin45## perpendicular to the ##F_{AP}##? F_BP sin 45 = 31.8N

I got the answer, and I edited my answer in post #1.