According to Coulomb's law, rank the interactions between charged particles from lowest potential energy to highest potential energy. a. a 1+ charge and a 1- charge separated by 100 pm b. a 2+ charge and a 1- charge separated by 100 pm c. a 1+ charge and a 1+ charge separated by 100 pm d. a 1+ charge and a 1- charge separated by 200 pm

Video transcript

hey everyone in this example we need to use columns law to identify the charged particle pair with the smallest potential energy. We should recall our formula for columns law Which states that our potential energy is found from taking our column constant which is represented by 1 4th pi times epsilon not. And then multiplied by our quotient. Where we have our first charge multiplied by the second charge divided by our radius which would be our distance between our ion centers of our charges. Now for this example, because we know this portion of our formula is a constant. We can go ahead and actually simplify our formula so that we just have energy equal to our first charge Q1 times our second charge Q2 divided by their radius. So let's go ahead and begin part A. In part they were given that we have our first charge Q1 equal to value of positive too. And then this is multiplied by the second charge given as -2. We're told that our radius Between these two charges is equal to km. And these units PICO meters are technically going to be canceled out because of our units associated with our columns constant. So we can go ahead and just plug in 200 in our denominator. And so this is going to give us a value equal to negative 0.02. So we're going to keep comparing and move on to example B An example be we're going to find the potential energy by taking in our numerator, the first given charge which is given as positive three and multiplying it by the second charge. Q two given as negative one. And then we're dividing it by their radius from one another given as 150 PK meters. So when we take the value of this quotient, we're going to get a difference equal to -0.02. So so far we have similar magnitudes in part A and B. Let's go ahead and consider part C. Where we're going to find the potential energy equal to in our numerator, the product of our first charge given as plus three. Multiplied by the product Or sorry, our second charge given as -2 and then divided by their distance from one another given as 250 km. So when we take the product of this quotient, we're going to get a value equal to negative 0.024. And so we would conclude conclude this problem by saying therefore Q one equal to positive three and Q one equal to negative two are the ion pair with the smallest potential energy. And so this would confirm that our correct choice for this question is going to be choice. E So this will complete this example. If you have any questions please leave them down below and I'll see everyone in the next practice video