Question #618a2
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"What is an example of an Ideal gas law practice problem?"
Their electrical potential energy is 22,500 J.
The electric potential energy associated with two charges, #q_1# and #q_2#, separated by a distance #r# can be written like this
#U = k_e * (q_1 * q_2)/r#, where
#k_e# - Coulomb's constant, equal to #9.0 * 10^(9)"N m"^2"C"^(-2)#.
As you can see, this is simply a plug-and-play problem in which you use the given values to determine #U#.
SInce one spere is 75 mm from point P and the other is 25 mm from the same point P, in opposite directions, it follows that the distance between them is equal to
#r = 75 + 25 = "100 mm"#
This means that their electric potential energy is
#U = k_e * (q * q)/r#
#U = 9.0 * 10^(9)"N" * "m"^(cancel(2))/cancel("C"^2) * ((500 * 10^(-6))^2cancel("C"^2))/(100 * 10^(-3)cancel("m")#
#U = 9.0 * cancel(10^9) * 250000/100 * cancel(10^(-9))underbrace("N" * "m")_(color(blue)("=Joule"))#
#U = color(green)("+22,500 J")#
