what is the magnitude of the electric field to the nearest tenth of a n/c

A field is a means of thinking about and visualizing the strength that surrounds any charged object and acts on another charged object at a distance, fifty-fifty if there is no obvious concrete contact between these two objects.

Electrical field $Due east$ due to gear up of charges at any point is the force experienced by a unit positive examination charge placed at that point.

The units of electric field are $N/C$ or $5/m$.

Electric field $E$ is a vector quantity meaning it has both

• magnitude and
• direction

In this commodity nosotros volition learn how to find the magnitude of an electric field.

Magnitude of electrical field created by a charge

Let us at present first expect at finding magnitude of electric field created by a charge.

To find electric field due to a unmarried charge we make utilize of Coulomb's Law.

If a betoken accuse $q'$ is at a distance $r$ from the accuse $q$ and then information technology will feel a forcefulness

$\vec{F}=\frac{one}{four\pi \epsilon_0}\frac{qq'\lid{r}}{r^ii}$

Electric field at this indicate is given by relation

$\vec{Due east}=\frac{\vec{F}}{q'}=\frac{1}{4\pi \epsilon_0}\frac{q\hat{r}}{r^2}$

This is electric field at a distance $r$ from a point accuse $q$ and $\hat{r}$ is the unit vector forth the direction of electric field.

Above relation defining electrical field at a distance $r$ tels about both magnitude and direction of the field.

Magnitude of electrical field would exist

$|\vec{E}|=\frac{1}{4\pi\epsilon_0}\frac{|q|}{r^2}$

In above equation you could notice the missing $\chapeau{r}$ part. This $\hat{r}$ tells us virtually the direction of the electrical field. Since electric force is a central strength and nosotros have divers electric field using Coulombs law nosotros can conclude that electric field acts forth the line joining the charge $q$ (source point) and field point at which it is beingness measured.

This vector This $\hat{r}$ is known as unit vector of our displacement vector $\vec{r}$ and is given by relation

$\chapeau{r}=\frac{\vec r}{r}$

Where $r=|\vec{r}|$ which is the altitude between our source signal and the field point.

Magnitude of electric field Formula

Thus, the magnitude of electric field due to a point charge is given past relation

$E=\frac{F}{q'} =\frac{ane}{4\pi\epsilon_0}\frac{q}{r^2}$

It is important to notation here that the magnitude of $\vec E$ depends on the charge $q$ wgich produces the electric field non on the value of test charge $q'$.

Solved Example 1

Find the magnitude of electric field at a distance $0.2m$ from a charge of $4nC$.

Solution

Step 1 – Figure out what is requited

The question asked the states to find magnitude of electric field

Step – 2 Figure out data given in the question.

We are provided the magnitude of the charge as well as the altitude betwixt the field indicate and the accuse.

Step iii- Find out the way to solve problem

We will use the relation

$Due east=k\frac{Q}{{r}^{ii}}$

To find the magnitude of electric field.

Step 4- Solve the trouble

$East=k\frac{Q}{{r}^{2}}$

where

$k=\frac{1}{four\pi\epsilon_0}=nine.0\times 10^nine North\cdot m^/C^2$

So,

$E=(nine.0\times 10^nine)\frac{4\times x^9}{{.2}^{2}}=900N/C$

Magnitude of electric field due to multiple charges

Magnitude of electric field due multiple charges tin can be calculated using superposition principle.

It states that "The full electrical field at a point P is the vector sum of the fields at P due to each point charge in the charge distribution."

To explicate this further let us consider the figure which shows two electrical charges $q_1$ and $q_2$ and we have to find net electric field at point P due to these ii charges.

Co-ordinate to principle of superposition

$\vec E = \frac{\vec F}{q'}=\vec{E_1}+\vec{E_2}$

If nosotros have knowledge about the magnitude of charges and altitude of point P from both these charges then we can use relation

$E=1000\frac{Q}{{r}^{2}}$

Where $k=\frac{one}{4\pi\epsilon_0}=9.0\times x^9 N\cdot m^/C^2$

To approximate the resultant field, we must calculate the electric field for each accuse separately and and then add together them together.

Let u.s. now understand this with the assist of a problem.

Solved Example 2

Two point charges $q_A=3\mu C$ and $q_B=-3\mu C$ are placed at xx cm apart in vacuum. What would be the electrical field at the mid point $O$ of the line $AB$ joining the two charges?

Solution

Pace- 1 Figure out what is requited

The question asked us to discover magnitude of electric field at the center of line joining both the charges.

Step- 2 Figure out information given in the question.

We are provided the magnitude of the charge as well as the altitude betwixt both the charges

\({q_A} = three\mu C = 3 \times {ten^{ – vi}}C\)

\({q_B} = -3\mu C = 3 \times {10^{ – 6}}C\)

\(r=20 cm = 0.2m\) we have to catechumen $r$ given in cm to m.

Allow $0$ exist the mid signal of the line equally shown below in the figure

So,

$OA=OB=\frac{r}{2}=\frac{0.ii}{two}=0.1m$

Step- 3 Find out the way to solve problem

We will employ the relation

$E=one thousand\frac{Q}{{r}^{two}}$

To find the magnitude of electric field.

Step 4- Solve the trouble

Kickoff we will calculate electric field at point O due to $q_A$

$E_A=k\frac{|q_A|}{(OA)^two}=9\times 10^ix\times \frac{3 \times {10^{-six}}}{(0.ane)^two}=2.7\times ten^six North/C$

Acting along OB

Side by side we will summate electric field at point O due to $q_B$

$E_B=k\frac{|q_B|}{(OB)^2}=9\times 10^9\times \frac{3 \times {10^{-6}}}{(0.1)^2}=2.7\times 10^6 North/C$

Interim along OB

Now since $E_A$ and $E_B$ are interim along same direction the angle between them is $\theta = 0\, radians$ and $coso = 1$

$|Due east| = \sqrt {E_A^2 + E_B^ii + ii{E_1}{E_2}cos\theta } $

$ = \sqrt {E_A^2 + E_B^two + 2{E_1}{E_2}} $

$ = |{E_A}| + |{E_B}|$

Therefore net magnitude of electric field at point O due to two charges $q_A$ and $q_B$ would be,

$Eastward=E_A+E_B=two.7\times ten^6+2.7\times 10^6 = 5.4\times 10^6 N/C$

Note:- If $E_A$ and $E_B$ acts along reverse direction the angle between them is $\theta = \pi \, radians$ and in that case $|E| = |{E_A}| – |{E_B}|$

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