Background 1. When an object has so much electric charge that the electric field
ID: 2059458 • Letter: B
Question
Background1. When an object has so much electric charge that the electric field at its surface is
greater than about 3 million N/C, the air around it suffers electrical breakdown, during
which the air temporarily acts like a conductor, allowing some of the charge to escape
in the form of a spark. If you've ever been shocked when reaching for metal doorknob,
you know what electrical breakdown is! Key point: no object can be charged past the
point where the air at its surface suffers electrical breakdown.
2. When an amount q of electric charge is distributed over the surface of a sphere, the
electric field outside the sphere is exactly the same as if the sphere were replaced by a
single point charge q where the sphere's center used to be.
3. If the above seems random and irrelevant, just wait.
Question
if you had two ping pong balls (or golf balls, or tennis balls), both carrying the maximum
possible positive electrical charge, and you also had insulating gloves that allowed you to pick
them up without being shocked, would you be strong enough to make the charged balls touch
each other?
Use background idea (1) above to figure out the maximum charge the ball of your choice can
hold. Remember that background idea (2) lets you treat a charged ball as if it were a point
charge (i.e., as if all its charge were concentrated at its own center), for the purposes of
calculating electrical fields and forces. You'll need to find out the size of the ball you choose,
and make a reasonable assumption about how much force you could exert on the balls in
trying to make them touch.
Explanation / Answer
The E field at the surface of one ball is 3,000,000 N/C. The charge on either ball is found by using E = kq/r^2 where k is a constant, q is the charge, and r is the radius of the ball. So... 3,000,000 = 8.99x10^9 * q / 0.02^2 where I estimated the radius of the ball as 2 cm q = 1.33 x 10^-7 Coulomb Now... the force required then is the E field from one ball times the charge from the other: F = qE = 1.33 x 10^-7 * 3 x 10^6 = 0.4 Newtons So sure... it would only require a force equivalent to about 0.4 N, which is a few ounces of force (i.e. a fraction of a pound.)
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