# Solving Electric Field Zero: 0.20m Charges

• jalen
In summary, the problem involves two charges of +1.5x10^-6C and +3.0x10^-6C separated by a distance of 0.20m, and the task is to find the point between them where the electric field is equal to zero. The attempt at a solution involves setting up an equation using the equation for electric field strength, but there seems to be an error in the calculations as the given answer of 1.2x10^-1m does not match with the calculated values. The poster is asking for assistance in identifying the incorrect line in their solution.
jalen

## Homework Statement

+1.5x10^-6C, +3.0x10^-6C, 0.20m

## Homework Equations

Two charges of +1.5x10^-6C and +3.0x10^-6C are 0.20m apart. Where is the electric field between them equal to zero?

## The Attempt at a Solution

= (9.0x10^9)(1.5x10^-6C) + (9.0x10^9)(3.0x10^-6C)
x^2 (0.2-x)^2
1.5(0.2-x)^2=3.0(x^2)
(0.2-x)^2=2(x^2)
0.4-0.2x-0.2x+x^2=2x^2
-x^2-0.4+0.04=0

The answers you'll get are -0.48 and 0.08. The answer in the text is 1.2x10^-1m and you can get it if it's 0.06 instead of 0.04. I looked over my work but I can't find where I went wrong. =(

The electric field strength due to Q1(at distance x from Q1)=The electric field strength due to Q2(at the distance of x from Q2)

Sorry, I don't really understand what you're trying to say. Can someone please tell me which line is incorrect so I can take a closer look and figure it out again.

## 1. How do you calculate the electric field at a point between two 0.20m charges?

To calculate the electric field at a point between two 0.20m charges, you can use the formula E = kq/r^2, where E is the electric field, k is the Coulomb's constant, q is the charge of each particle, and r is the distance between the particles. Plug in the values for these variables to find the electric field at the desired point.

## 2. Can the electric field at a point be zero if there are two 0.20m charges present?

Yes, the electric field at a point can be zero if there are two 0.20m charges present. This occurs when the two charges are equal in magnitude and opposite in direction, cancelling out each other's electric fields at that point.

## 3. How does the distance between two 0.20m charges affect the electric field at a point?

The electric field at a point between two 0.20m charges is inversely proportional to the square of the distance between the charges. This means that as the distance increases, the electric field decreases, and vice versa.

## 4. What units are used to measure the electric field?

The electric field is measured in units of Newtons per Coulomb (N/C) in the SI system. In the CGS system, the unit is statvolts per centimeter (statV/cm). Both of these units represent the force per unit charge at a point in an electric field.

## 5. How can you use the electric field equation to solve for the charges if the field is known?

If the electric field at a point is known, the equation E = kq/r^2 can be rearranged to solve for the charge q. First, multiply both sides by r^2, then divide by k. This will give you the equation q = E*r^2/k. Plug in the values for E and r to find the magnitude of the charge at that point.

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