For this problem on the topic of electric potential, we have two point charges each with the magnitude to micro columns that are located on the X axis. One is that X is equal to one m and the other X is equal to minus one m. We want to find the electric potential on the Y axis at Y. Is equal to 0.5 m. We then want to calculate the electric potential energy of a third charge of magnitude minus the micro columns. It is placed on the Y axis at Y is equal to 0.5 m. So firstly the potential V at our point label, P is simply the potential due to charge Q. One which is K E Q one over our one must be potential due to charge too, which is K E E Q two over our two. And this is equal to two into K E Q. Of our. And since the charges that is equal between the two as well as their separation. And so this is to into eight 0.99 times 10 to the nine that's K. Newton meters squared, McCullum squared, multiplied by the magnitude of the charge, which is to times 10 to the minus six columns, all divided by their distance to point P, which is the square root of one meter squared plus zero point five m squared. And so calculating, we get the potential any point that is 0.5 m along the y axis to be three 0.22 times 10 to the four vaults, which is 32 going to kilovolts. Next, we want to find the electric potential energy of a third charge that is placed on the y axis at Y. Is equal to 0.5 m. So the electric potential energy, you is able to cue times V. This is equal to the charge minus three times 10 to the minus six columns, plus the potential there, which is three. What time's the potential? Rather, the 0.22 times 10 to the power four jules, A colon. This gives the electric potential energy at this point to be minus nine 0.65 times 10 to the minus two jules.