This is a problem on electric current

[heavenaintclo]

Homework Statement

An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a current of 1.000 000 0 A out of it. How long would it take for the sphere to increase in potential by 1200 V?

Homework Equations

n/a

The Attempt at a Solution

tried using i= dq/dt

 

[m.e.t.a.]

You'll need to rearrange the following equation to find how much charge, $q$, the sphere will store when its surface potential is at 1,200 V.

$$V_R=\frac{q}{4 \pi \epsilon_0 R^2}$$

You know the magnitudes of the inward- and outward-flowing currents, therefore you can calculate the net inward-flowing current. All you need to do next is work out how long it takes for this current to transfer charge $q$.

 

[nlightner]

and q=CV but couldn't find a way to get time.

I would approach this problem by first identifying all the given parameters and equations that can be used to solve it. In this case, we have the radius of the conducting sphere, the currents in the wires, and the desired change in potential. The equation i = dq/dt can be used to relate the current to the rate of change of charge, and q = CV can be used to relate charge to potential difference, where C is the capacitance of the conducting sphere.

To solve for time, we can use the equation t = q/i, where t is the time, q is the change in charge, and i is the current. Using the given values, we have q = CV = 1200 V x (4πε0r^2)/10 cm = 4.8πε0r^2 C, where ε0 is the permittivity of free space. Plugging this into the equation for time, we have t = (4.8πε0r^2 C)/(1.000 002 0 A - 1.000 000 0 A) = 1.2 x 10^-6 seconds.

Therefore, it would take approximately 1.2 microseconds for the isolated conducting sphere to increase in potential by 1200 V. This solution assumes ideal conditions and neglects any resistance in the wires or other factors that may affect the actual time. Further analysis and experimentation may be necessary to obtain a more accurate result.