Question

two positive charges of 1 microcoulomb and 2 microcoulomb are placed at 1 m apart the value of electric field in Newton per full at the middle point of the line joining of the charges will be

Answer 1

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Two positive charges of 1 microcoulomb and 2 microcoulomb are placed at 1 m apart the value of electric field in Newton per full at the middle point of the line joining of the charges will be

Note: the electric field at the midpoint due to both charges act in opposite directions, then: d² => (d/2)²

1 μC ._______Q1> __ <Q2_________. 2 μC

______________ 1 m ______________

Solving:

[First Step] Conversion of μC (microCoulomb) into C (Coulomb)  

Then the two punctual or punctiform charges q1 and q2 will look like this:

$Q_1 = 1\:\mu{C}\to Q_1 = 0.000001\:C\to\:\underline{Q_1 = 1*10^{-6}\:C}$

$Q_2 = 2\:\mu{C}\to Q_2 = 0.000002\:C\to\:\underline{Q_2 = 2*10^{-6}\:C}$

[Second Step] We have the following data, to E1 :

$E_1\:(electric\:field\:strength) =\:?(in\:N/C)$

$k\:(electrostatic\:constant) = 9*10^9\:N*m^2*C^{-2}$

$Q_1\:(punctiform\:charge) = 1*10^{-6}\:C$

$d\:(source\:load\:distance) = 1\:m$

We will apply the data to the electric field intensity formula, see:

* to Q1

$E_1 = k* \dfrac{Q_1}{d^2}$

$E_1 = 9*10^9* \dfrac{1*10^{-6}}{\left(\dfrac{d}{2}\right) ^2}$

$E_1 = 9*10^9* \dfrac{1*10^{-6}}{\dfrac{1^2}{4}}$

$E_1 = \dfrac{9*10^3}{\dfrac{1}{4}}$

$E_1 = 9*10^3*\dfrac{4}{1}$

$E_1 = 36*10^{3}}$

$\boxed{E_1 = 3.6*10^{4}\:N/C}}$

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[Third Step] We have the following data, to E2 :

$E_2\:(electric\:field\:strength) =\:?(in\:N/C)$

$k\:(electrostatic\:constant) = 9*10^9\:N*m^2*C^{-2}$

$Q_2\:(punctiform\:charge) = 2*10^{-6}\:C$

$d\:(source\:load\:distance) = 1\:m$

Note: the electric field at the midpoint due to both charges act in opposite directions, then: d² => (d/2)²

We will apply the data to the electric field intensity formula, see:

* to Q2

$E_2 = k* \dfrac{Q_2}{d^2}$

$E_2 = 9*10^9* \dfrac{2*10^{-6}}{\left(\dfrac{d}{2}\right) ^2}$

$E_2 = 9*10^9* \dfrac{2*10^{-6}}{\dfrac{1^2}{4}}$

$E_2 = \dfrac{18*10^3}{\dfrac{1}{4}}$

$E_2 = 18*10^3*\dfrac{4}{1}$

$E_2 = 72*10^{3}}$

$\boxed{E_1 = 7.2*10^{4}\:N/C}}$

[Fourth Step] Therefore, the Net Electric Field will be:

$E_{Net} = E_2-E_1$

$E_{Net} = 7.2*10^4 - 3.6*10^4$

$\boxed{\boxed{E_{Net} = 3.6*10^4\:N/C}}\:\:\:\:\:\:\bf\purple{\checkmark}$

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$\bf\purple{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

Answer 2

Explanation:

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