My solution so far:
E = kQ/r2
E = kQ/a2 (^r)
I think tan(45*) should be used, so it should be...
E = kQ/a2 (tan(45*))
Thank you.
Forces due to two positive q charges cancel out each other. The remaining force is due to +q and-q which are at opposite corners.These two forces add up.
Final F= 2*kqQ/a2
Direction is along the diagonal from +q towards -q charge along the arrow drawn in the figure.
My solution so far: E = kQ/r2 E = kQ/a2 (^r) I think tan(45*) should be...
4. 15 points KatzPSEf1 23.P.051 My Notes Four equally charged particles with charge q are placed at the corners of a square with side length L, as shown in the figure below. A fifth charged particle with charge Q is placed at the center of the square so that the entire system of charges is in static equilibrium. What are the magnitude and sign of the charge Q? (Use any variable or symbol stated above as necessary.) magnitudeQ sign --Select...
Consider the symmetrically arranged charges in the figure, in which q a = q b = − 1.55 μC and q c = q d = + 1.55 μC . Four charges located at the corners of a square. Charge q subscript a is at the upper left corner. Charge q subscript b is at the upper right corner. Charge q subscript c is at the lower left corner. Charge q subscript d is at the lower right corner. A...
Please help, this is as far as I can get without getting
stuck! Thank you in advance!
PHYS 121 Quiz 2 Three and th charges are located at the corners of an equilateral triangle as shown. Find the direction e magnitude of the net electric field at the center of the equilateral triangle. Show the electric field in the picture, show your calculation for finding the magnitude and direction. Ei
1)Consider the symmetrically arranged charges in the figure, in which ?a=?b=−2.05 μC and ?c=?d=+2.05 μC . Four charges located at the corners of a square. Charge q subscript a is at the upper left corner. Charge q subscript b is at the upper right corner. Charge q subscript c is at the lower left corner. Charge q subscript d is at the lower right corner. A fifth charge q is located at the exact center of the square. Determine the...
Basic Coulomb's law tells us that the electric field of a point
charge q, E=ke q/r2, scales with distance r
as 1/r2: when the distance increases by a factor of 2,
the magnitude of the field decreases by a factor of 4. We can write
this as E(r) ∝ 1/r2. An interesting and practically
important question is how the electric field of a group of charges
scales with distance (asymptotically) far away from the group. The
figure below depicts 3...
Help with physics please
Coulomb constant is 9 × 109 N·m2/C2 3, 135° 4. 90° 5.315° < θ < 360° 6.225° 7.0。 8. 45 <90° θ< 180° Answer in units of N θ<135° 021 (part 1 of 2) 10.0 points Consider a square with side a. Four charges +9, +9-g, and +q are placed at the corners A, B, C, and D, respectively θ < 270° θ<45° 023 10.0 points Two identical small charged spheres hang in equilibrium with equal...
The charge for Q3 is not given. I could not find formula in
book. So do I take EA total 4.5 X 10^6 N/C. And use it for E? Then
set formula like so 4.5 X 10^6= (9.0x10^9)Q/(.30m)^2 and solve for
Q to get charge? I need help.
ple 16-9, but calculate the electric field (magnitude and direction) at Q1 due to the Rework Exam other two charges. EXAMPLE 16-9 E above two point charges. Calculate the total ele field...
Please explain how you are achieving these answers so that I can
figure out how to do this on my own as well
Part a In cases A and B shown in Figure 3 there are two positive point charges +Q each a distance s away from a third positive point charge +q Is the net electric force on the +q charge in case A greater than, less than, or equal to the net electric force on the +q charge...
Problem A2 - Understand Coulomb's Law 12 Write down the vector form of Coulomb's Law for the force on a point charge q, which is a distance r from a point charge 42- Sketch and clearly indicate the starting point and direction of all vectors involved. Problem A3 - The connection between Electric field and Coulomb's Law f3) The earth has a net electric charge that causes at points near its surface an electric field equal to 150 N/C and...
in Part c) I am not sure how they came up with 0.5 for
r for E1. will you please give a clear details as you solve it, and
please just go ahead and solve 47 so I can check my answers. Thank
you.
Step 6 of 8 A Electric field due to first charge is E,卡 kq Substitute 9x10, N-m2 /C2 for k, 6.00x10" С for q and 0.5 m for 5 , (9x10, N-㎡/C")(6.00x10° c) (0.5 m) E,...