Three negative point charges q1 =--5 nc, q2 = -2 nc and q3 = -5 nc lie along a vertical line. the charge q2 lies exactly between charge q1 and q3 which are 16 cm apart. find the magnitude and direction of the electric field this combination of charges produces at a point p at a distance of 6 cm from the q2.

Ep= -10400 N/C

Ep= 10400 N/C  In the direction of the (-x )axis

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence  

1nC= 10⁻⁹C

1cm = 10⁻²m

Data

K= 9x10⁹N*m²/C²

q₁ = q₃= -5 nC= -5 *10⁻⁹C

q₂ = -2 nC = -2 *10⁻⁹C

d₁=d₃=d

=  10*10⁻²m = 0.1m

d₂ = 6cm =  6*10⁻²m = 0.06m

Graphic attached

The attached graph shows the field due to the charges:

E₁ : Electric Field at point P  due to charge q₁. As the charge q₁  is negative (q₁-) ,the field enters the charge

E₂: Electric Field at point  P  due to charge q₂. As the charge q₂  is negative (q₂-) ,the field enters the charge

E₃: Electric Field at point  P  due to charge q₃. As the charge q₃  is negative (q₃-) ,the field enters the charge

Ep: Total field at point P due to charges q₁ , q₂ and q₃ .

Because q₁ = q₃ and d₁ = d₃, then, the magnitude of E₁ is equal to the magnitude of E₃

E₁ = E₃ = k*q/d² = ( 9*10⁹*5*10⁻⁹) /(0.1)² = 4500 N/C

E₂=  k*q₂/d₂²  = ( 9*10⁹*2*10⁻⁹) /(0.06)² = 5000 N/C

Look at the attached graphic :

Epx : Electric field component at point P in x direction

Epx =E₁x+E₂x+ E₃x

E₁x= E₃x= - 4500 * cosβ =- 4500* (6/10) = - 2700 N/C

E₂x = - 5000 N/C

Epx = E₁x+E₂x+ E₃x = - 2700 N/C - 5000 N/C- 2700 N/C

Epx = - 10400 N/C

Epy : Electric field component at point P in y direction

Epy =E₁y+E₂y+ E₃y

E₁y=  4500 * sinβ = 4500* (8/10) =  3600 N/C

E₃y= - 4500 * sinβ =  -4500* (8/10) =  -3600 N/C

E₂y = 0

Epy =E₁y+E₂y+ E₃y = 3600 N/C+0-3600 N/C

Epy = 0

Magnitude and direction of the electric field this combination of charges produces at a point P

Ep= Epx = 10400 N/C  In the direction of the (-x) axis

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

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so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards