Recent content by Antonius

So, I can just use B = μI/2πr, where r is distance from wire 1 to wire 2, to find the Magnetic field that ## I_1 ## creating? Can you shortly explain the reason behind not integrating to find ## B ##? I am not quite clear on that...

Homework Statement Homework Equations Biot-Savart Law: ## dB = μIdl/4πr^2 F = ILxB The Attempt at a Solution I have not tried to solve it. BUT, please check my approach. I want to make sure my method is correct and whether there is a flaw or no. I am trying to set up an integral here. I am...

The question: Parametrize the solution set of this one-equation system. x_1 + x_2 + ... + x_n = 0 My question (please look at the photo): I understood why we have the first row, but what's the point of the other rows?

oh

There is a vague step which I did not fully understand at first due to weak math background. Use "the definition of derivative" and it will be much clear and you will most likely get a correct result

Homework Statement Suppose you have two identical droplets of water, each carrying charge 7.92 pC spread uniformly through their volume. The potential on the surface of each is 144 Volts. Now, you merge the two drops, forming one spherical droplet of water. If no charge is lost, find the...

I got -2.15e5. Is this what you also got?

I did it before subbing a number for r though? And r is not 0.4168 m it's 0.0458 m ( in the question it says its r = 4.58 cm )

If ## V (at 9.1r) = \frac {kq} {9.1*r} ## then ## \frac {dV} {dr} = \frac {kq} {9.1} *(\frac {-1} {r^2}) ## here ## q = \frac {V_r * r} {k} ## which I plugged into the equation and got ## \frac {dV} {dr} = \frac { -V_r} {9.1*r} ## which makes ## \frac {dV} {dr} = -23635.97 = -2.36e10^4 ##...

Hmm. What's going on lol. Okay, let me work on it again. so if q = 5.069 e -8 Now I need to find V at distance 9.1*r (away from center of sphere) So V = kq/9.1*r = k*5.069 e -8 / (9.1*0.0458) = 1082.44 Volts Potential Gradient = delta V / delta x = (1082.44 V - 9851 V) / (9.1*r - r)...

Yes, I see. I wrote -7, instead of -8. The final answer that I have gotten is calculated with 5.069*10^(-8). Thank you!

Apparently, I forgot to put a parenthesis while dividing by 8.9e9. jtbell, thank you! So, q = 5.07e-7 Hence, ## \frac {\Delta V} {\Delta r} = -2.36e4 V/m ## I am hoping this answer is correct.

So I did this: $$ V = \frac {kq} {r}; 9851 Volts = \frac { 8.9 * 10^9 * q } { 0.0458 m } ; Hence, q = 5.069 C $$ Now since I know the ## q ## I decided to find ## V ## at ## 9.1 r ##. To do so I followed these steps: $$ V = \frac { 8.9 * 10^9 * 5.069 } { 9.1 * 0.0458 } = 1.08 * 10^11 V $$...

[Note from mentor: This was originally posted in a non-homework forum so it doesn't have the homework template.] ----------------------------------------- Problem: The surface of a solid metal sphere (radius r = 4.58 cm) is at potential V = 9,851 Volts. Find the magnitude of the potential...

Oh I see,