Question

two point charges each of 5 micro Coulomb but opposite in sign are placed 4cm apart calculate electric field at a point distance 4cm from of axial point
​

Answer 1

Answer:

The electric field intensity is 0.562\times10^{8}\ N0.562×10

8

N .

Explanation:

Given that,

Charge q=5\times10^{-6}\ coulombq=5×10

−6

coulomb

Distance r= 4\times10^{-2}\ meterr=4×10

−2

meter

The electric field intensity at a point 4 cm away from the midpoint of the axial line is

The electric field intensity on the axial line define as:

E_{axial}=\dfrac{k2p}{r^3}E

axial

=

r

3

k2p

Here, p = ql

E_{axial}=\dfrac{9\times10^{9}\times2\times5\times10^{-6}\times4\times10^{-2}}{(4\times10^{-2})^3}E

axial

=

(4×10

−2

)

3

9×10

9

×2×5×10

−6

×4×10

−2

E_{axial}=0.562\times10^{8}\ NE

axial

=0.562×10

8

N

Hence, The electric field intensity is 0.562\times10^{8}\ N0.562×10

8

N

Hope it helps☺

Thank you