Question

Two point charges, q1 = 2.0 × 10−7 C and q2 = −6.0 × 10−8 C, are held 25.0 cm apart. (a) What is the electric field at a point 5.0 cm from the negative charge and along the line between the two charges? (b)What is the force on an electron placed at that point?

Answer 1

Answer:

a)$432000\frac{N}{C}\hat{i}$

b)$-6.92\times10^{-14}N\hat{i}$

Explanation:

a)

The magnitude of the electric field generated by a charged particle at a distance r is:

$E= k\frac{|Q|}{r^{2}}$

With Q the charge of the particle and k the constant ()

So, the electric field generated by q1 knowing that the point 5.0 cm apart the negative charge is $25.0cm-5.0cm=20.0 cm=0.2m$ apart the positive charge is:

$E_1= k\frac{|q1|}{r_1^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|2.0\times10^{-7}|}{(0.2)^{2}}$

$E_1= 45000\frac{N}{C}$

and the electric field generated by q2:

$E_2= k\frac{|q2|}{r_2^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|-6.0\times10^{-8}|}{(0.05)^{2}}$

$E_2=216000\frac{N}{C}$

Those are the magnitudes of the electric field, but electric field is a vector quantity, so the direction is important. Electric field generated by negative particles points towards the charge and electric field generated by positive particles points away the particle. So, if we define positive direction towards negative particle (x-axis):

$\overrightarrow{E_2}=+216000\frac{N}{C}\hat{i}$

$\overrightarrow{E_1}= +45000\frac{N}{C}\hat{i}$

Vector quantities satisfy superposition principle, this is $\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$, with E the total electric field.

$\overrightarrow{E}=(216000+45000)\hat{i}=432000\frac{N}{C}\hat{i}$

b) The force is:

$\overrightarrow{F}=e*\overrightarrow{E}$,

with q the charge of an electron

$\overrightarrow{F}=(-1.61\times10^{-19})*(432000)\hat{i}=-6.92\times10^{-14}N\hat{i}$