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Answer #1

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Question 1 :

Given, total charge on the rod (Q) = 8.30 nC

length of the rod (L) = 1.40 cm

distance of point P from the center of the rod (l) =8.80 cm = 8.80 x 10-2 m

So, charge per unit length on the rod (ho) is

8.30 x 10- 1.40× 10-2 = 5.93 × 10-7 C /m

丼 1

I have taken a charge element dx at a distance x from the origin O on the rod.

Distance of the charge element from the point P is , r = ( l + L/2 - x ) .

Charge on the element dx = ho , dx

So, electric field due to this element at point P is

  dE = rac{1}{4 pi epsilon _{0}} rac{ ho , dx}{r^{2}} = rac{1}{4 pi epsilon _{0}} rac{ ho , dx}{left ( l + rac{L}{2} - x ight )^{2}}

Where,

rac{1}{4 pi epsilon _{0}} = 9 imes 10^{9} , , Nm^{2}/C^{2}    is constant .

and ho is also a constant .

So,   da 0 ..............................(1)

NOTE :

This type of integration can be done as follow :

d.x (a -z)2

(a-x)-2+1 = 2 (a   

Now from equation (1), we get

E= rac{ ho}{4 pi epsilon _{0}} left [ rac{ 1}{left ( l + rac{L}{2} - x ight )} ight ]_{0}^{L}

  = rac{ ho}{4 pi epsilon _{0}} left [ rac{ 1}{left ( l + rac{L}{2} - L ight )} - rac{ 1}{left ( l + rac{L}{2} - 0 ight )} ight ]

= rac{ ho}{4 pi epsilon _{0}} left [ rac{ 1}{left ( l - rac{L}{2} ight )} - rac{ 1}{left ( l + rac{L}{2} ight )} ight ]

  = rac{ ho}{4 pi epsilon _{0}} left [ rac{ L}{left ( l ^{2} - left ( rac{L}{2} ight )^{2} ight )} ight ]

putting all the values, we get

E=9×109 × 5.93 × 10-7 × 1.40 x 10-2 (8.8 × 10-2)2-( 1.40×10-2 ) 2

  74.7 76.95 × 10-1 9.7 × 10-M/C

And the direction of this field will always along the positive X-axis.

Question 2 :

This is simply a case of projectile motion . Here the acting force is electrostatic instead of gravitational.

We know the equation for range (R) in projectile motion (in case of gravitation) :

u2 sin 2θ

here, g is the acceleration due to gravity which always acts downward.

u is the initial velocity of the projectile and

heta is the launch angle of projectile with x - axis in anticlockwise direction

Comparing this situation with our problem.

Here, our projectile is electron. The positive plate of capacitor will always attract the electron towards it .

Consider the positive plate as x- axis and a line from positive plate to negative plate as y -axis.

So, instead of g (acceleration due to gravity) here the acceleration of the electron will be

de me (from Newton's second law of motion )

Where, Fe is electrostatic force and  

Fe = eE (E is electric field)

So, de me

So, we can write equation for range in this case as

u2 R- sin 20

Given,

R = 2.50 cm = 2.50 x 10-2 m

u = 2.45 x 106 m/s

θ = 45°

Putting all the values in equation for R, we get

(2.45 × 106)2 × sin (2x 45°) -2 2.50 × 10-2- Cle

or, (60025 × 1012) 2.5 × 10-2 Cl

or, a.-2.4 x 1014 m/s

We have,

de me

Where, e = 1.602 x 10-19 C is electronic charge

me = 9.1 x 10-31 kg is mass of electron

So, 2.4 × 10x 10-19 × E 14 1.602 9.1 × 10-31

or, 2.4 × 1014-0. 176 × 1012 × E

Or, E= 13.63× 102 N/C

This is strength of electric field between the plates of the capacitor.

For any doubt please comment and please give an up vote. Thank you.

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