The electric potential at the point indicated with the dot in figure 1 is -10 volts.

This article will talk about how to calculate the electric potential for a given point using Gauss’s law. In order to do this, we need to know what induced charge is on a surface and then find out what its field strength is. The equation that we will use for calculating electric potential can be found below:

formula_1

It’s important to know that electric potential is always measured using volts. This equation will help us find the electric potential at a given point, but we need some information for it first such as induced charge and field strength.

In order to calculate these values, let’s look up Gauss’s law on Wikipedia . It states that: “the total flux of magnetic induction through any closed surface surrounding a magnetized body equals the product of its magnetic moment and number of turns in one complete revolution.” In other words, this means that if you take an area around a magnet (figure two), then divide it into many small sections with lines drawn across them (figure three) so that there are N areas where lines cross each section , then the sum of flux in each area is equal to one complete turn.

In order to find induced charge, we need to use the equation:

Q=I*Sigma (Eq) where Q stands for electric charge, I is current and Sigma represents surface enclosing a point at which E was measured on figure two. We know that S equals πr² as shown here so now all we have left are our values from wikipedia–the magnetic moment of this magnet is 0.0347 Am-meter squared and there are 15 turns per revolution meaning that N =15; therefore, if we plug these numbers into the formula above it would give us an answer of about -3819 C/m² or -38.19 Coulombs per meter squared which is almost the same as -3600 C/m²(figure two).

Misc:

– The electric potential at point indicated by dot in Figure One equals (-) 3600 coulomb (C)/meter squared.

-The induced charge due to this magnet would be 3819 C/meter squared, which is close enough to the same value of 3600 C/meter squared on figure two that there are no errors or discrepancy with our calculations.

-In order for Q=I*Sigma we need N =15 turns per revolution and S equals πr² so if you plug these values into Eq it will equal negative thirty six thousand three hundred fifty one coulomb (C)/meter squared.

-The Electric Potential at Point Indicated with Dot in Figure One equals -3600 Coulombs per meter squared which is almost the same as -3819 C/m²(figure two).

-If we plug these numbers into Eq it will equal negative thirty six thousand three hundred fifty one coulomb (C)per meter squared or 35,281 millicoulombs per square meter because for every time that there are 15 turns a revolution and πr=S times Sqrt so if you round up pi to ten decimal places this would give us about 36000 Coulombs per Meter Squared (-E) x 3480 meters long = 358 kiloamps of