# What is the magnitude of the electric field

• xswtxoj
In summary, the problem involves finding the magnitude of the electric field at a specific point (x=0.506 m, y=0.506 m) between two charges (+21.3 nC and +11.0 nC) located at different points along the x-axis. To solve this, one must first calculate the individual electric field magnitudes of each charge based on their distances from the point. Then, the magnitudes must be separated into their x and y components and added together to determine the total electric field magnitude at the point. Finally, the result can be found using the Pythagorean theorem.
xswtxoj

## Homework Statement

Two electric charges, q1 = +21.3 nC and q2 = +11.0 nC, are located on the x-axis at x
= 0 m and x = 1.00 m, respectively. What is the magnitude of the electric field at the
point x = 0.506 m, y = 0.506 m?

How would you think to approach the problem?

21.3+11.0 /1 = .506/.506 ?

Based on what?

They want you to find the E-field vector at .506,.506.

First calculate the |E1| and |E2| based on the geometric distance to both charges.

Then separate those vectors - they are vectors - into their components.

Then add the x-components and y components separately.

Happily they only want the magnitude of the E-field - |E| - so just use Pythagoras to figure the |E| of the resulting vector.

theres no angle though, so it would be .506 sin____ and .506cos___ or is it 90 or 180 degrees

xswtxoj said:
theres no angle though, so it would be .506 sin____ and .506cos___ or is it 90 or 180 degrees

You have the coordinates of the point.

and you know - or should know that a2 + b2 = c2

So ... figure it out.

.506 sq + .506 sq = 574.69 then sq root = .715,

xswtxoj said:
.506 sq + .506 sq = 574.69 then sq root = .715,

It is .715 but your intermediate result is nonsense.

So that's the r for 1 charge.

Now figure the r for the other.

21.3 sq + 11 sq= 574.7 equals 23.97 after sq rt, then take the 2 r's and add them up and tan y/x?

We're looking at distances, not charges.

would it be : E1= ke q1/ r sq? then E2= ke q2/r sq but how would the 2nd r be found would the x= 0, and 1 or 1, 0.506 since it for q2?

They want you to find the distance from 1,0 to .506,.506 for the second charge. That triangle then is .494,.506.

once i get r, i solve for e, then once i get both e's do i add them up or leave it as 2 separate answers?

xswtxoj said:
once i get r, i solve for e, then once i get both e's do i add them up or leave it as 2 separate answers?

Once you get the |E| (magnitude of E) then you must resolve them into their x,y components and add the components - being careful of the signs - and then determine the magnitude of the result.

They want the magnitude of the Total E field at that point.

## 1. What is the definition of electric field magnitude?

The magnitude of the electric field is a measure of the strength of the electric field at a particular point in space. It is defined as the force per unit charge at that point.

## 2. How is the magnitude of the electric field calculated?

The magnitude of the electric field is calculated by dividing the force on a test charge by the magnitude of the test charge. This can be represented mathematically as E = F/q, where E is the electric field magnitude, F is the force, and q is the test charge.

## 3. What is the unit of measurement for electric field magnitude?

The unit of measurement for electric field magnitude is newtons per coulomb (N/C) in the SI system. In other systems, it may be expressed as volts per meter (V/m) or dynes per statcoulomb (dyn/statC).

## 4. How does the distance from a source affect the magnitude of the electric field?

The magnitude of the electric field decreases as the distance from the source increases. This relationship follows the inverse square law, meaning that the magnitude of the electric field is inversely proportional to the square of the distance from the source.

## 5. Can the magnitude of the electric field be negative?

Yes, the magnitude of the electric field can be negative. This indicates the direction of the electric field, with a negative magnitude indicating that the electric field is pointing in the opposite direction of a positive test charge.

Replies
23
Views
3K
Replies
2
Views
7K
Replies
5
Views
1K
Replies
12
Views
2K
Replies
3
Views
2K
Replies
17
Views
2K
Replies
6
Views
1K
Replies
4
Views
2K
Replies
7
Views
2K
Replies
14
Views
4K