Recent content by miyayeah

Homework Statement Referring to (b) in the picture above: What is the direction of the magnetic force on the electron? Homework Equations I use the right-hand rule where by putting your index finger in the direction of the magnetic field and thumb in the direction of the velocity, the middle...

Homework Statement In the attached file, how do I decide which side of the resistor is positive/negative? Homework Equations N/A The Attempt at a Solution The current would go through the 6V battery starting from the positive end of the 9V battery to the negative end of the 9V battery; so my...

Thank you for pointing that out. I edited it now. If I were to sketch, I would draw the electric potential increasing from the negative to the positive end, and the electric field vector going from the positive to the negative end. I think I am mainly confused about where I should put the...

Homework Statement The electric field inside a parallel plate capacitor is measured to be E= -3500 N/C i. The electric potential at point XA = 3.00 m is measured to be 1500V. What is the electric potential at point XB = 0 m? Homework Equations V=E⋅s The Attempt at a Solution I think I need to...

I believe 4Ω and 48Ω resistors are in parallel. Req = (1/R1 + 1/R2 +...) -1. But how would this help solve the problem? I tried drawing it in two different parts, but I'm not sure how to combine the two, if they are right.

Homework Statement Find the current and potential difference through 3Ω resistor. Homework Equations ε - I1 (Ω) - I2 (Ω) ... = 0 The Attempt at a Solution I only got as far as finding the equation: ε - I1 (3Ω) - I2 (16Ω) = 0 (For complete loop around circuit) I know I have to find I1, but I...

Thank you for your explanation. Which formula should I use if I am solving for the electric field due to finite line of charge?

Homework Statement The problem I encountered goes like this: A 10cm long thin glass rod uniformly charged to 10nC and a 10cm long thin plastic rod uniformly charged to -10nC are placed side by side 4cm apart. What are the electric field strengths E1 to E3 at distances 1cm, 2cm, and 3cm from the...

To find the x component, E=[(8.99⋅109Nm2/C2)(6⋅10-9C)(0.447)] / (√0.052+0.12m)2 = 1928.8944 N/C To find the y component, E=[(8.99⋅109Nm2/C2)(6⋅10-9C)(0.894)] / (√0.052+0.12m)2 = 3857.7888 N/C However, despite using these values, I still get the same answer as my first attempt: xnet= 10788 N/C...

The electric field vector of the positive charge only has an x component, which I already have calculated. However I am not quite sure how I would find the x and y components of the electric field from the negative charge if I do not have any information about an angle in the first place.

Homework Statement The diagram is attached. What is the strength of the electric field at the position indicated by the dot in Figure 1? What is the direction of the electric field at the position? Specify the direction as an angle measured clockwise from the positive x axis. Homework...

Homework Statement A -12nC charge is located at (x,y) = (1.0cm, 0cm). What are the electric fields at the positions (x,y) = (5.0cm, 0cm), (-5.0cm, 0cm), and (0cm, 5.0cm)? Write each electric field vector in component form. Homework Equations E=k(q/r2) The Attempt at a Solution I was able to...

I am not sure, would the distance from the half point to the top simply be 125.66/4? If that is right, then energy lost = 62.83 (approximately) ?

How far the car travel would be the circumference, so : C=2πr = 2⋅π⋅20 = 125.66 (approximately)

The second equation shows that, but what I did was (before using the second equation) I tried to find the curved distance of the path from the middle point to the top point because the question indicates "2.0J per meter of track". By finding the distance traveled I can multiply that by 2.0J to...