Two large, parallel conducting plates carrying op­posite charges of equal magnitude are separated by 2.20 cm. (a) If the surface charge density for each plate has magnitude 47.0 nC/m^2, what is the magnitude of E in the region between the plates?

Video transcript

Welcome back everybody. We are taking a look at two equal and large metallic sheets placed across from one another with opposite charges. We are told that the distance between the two sheets is 54 mm and that the surface charge dead on each sheet is this value right here. 20.2 nano columns per meter squared, or 20.2 times 10 to the negative ninth columns per meter squared. We are tasked with finding what is the magnitude of the electric field between these two metallic sheets. Really? This this capacitor, right? So we have that our magnitude of the electric field is equal to the potential divided by the distance between them or the charge density divided by our electric constant. Now we have both of these terms. So we'll go ahead and use this formula right here. So we have that E. Is equal to the charge density divided by the electric constant. So let's go ahead and plug in our values. We have 20.2 times 10 to the negative ninth, divided by 8.85 times 10 to the negative 12th. Giving us a magnitude of the electric field to be 2.28 times to the third. Newtons per Coolum corresponding to our answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one