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Electric Charge and Field

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3 months ago

Electric Charge and Field Class 12th

In the following unclear reaction,

D $\overset{α}{\to} D_{1} \overset{β^{-}}{\to} D_{2} \overset{α}{\to} D_{3} \overset{γ}{\to} D_{4}$

Mass number of D is 182 and atomic number is 74. Mass number and atomic number of D₄ respectively will be________

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alok kumar singh

Contributor-Level 10

Please find the solution below:

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New answer posted

3 months ago

Electric Charge and Field Class 12th

Two point charges A and B of magnitude + 8 * 10^-6 and -8 * 10^-6C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 * 10⁴ NC^-1. The distance 'd' between the point charges A and B is:

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alok kumar singh

Contributor-Level 10

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{- 6}]}{d^{2}} * 4$

E₂ (done to q₂) = $\frac{k q_{2}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{- 6}]}{d^{2}} * 4$

Enet = E₁ + E₂

$= 2 [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{- 6}}{d^{2}}]$

Given Enet = 6.4 * 104

$\Rightarrow 2 * [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{- 6}}{d^{2}}] = 6.4 * 1 0^{4}$

d² = 9 * 10^-6 + 6

d= 3m

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New answer posted

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Electric Charge and Field Class 12th

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alok kumar singh

Contributor-Level 10

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

E₂ (done to q₂) = $\frac{k ? q_{2}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

Enet = E₁+ E₂

$= 2 [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}]$

Given Enet = 6.4 * 104

$? 2 * [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}] = 6.4 * 1 0^{4}$

d² = 9 * 10^-6 + 6

d= 3m

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