two fixed charges a and b of 5/10^6C each are reparated by distance 6 m.C is the midpoint of line joining a and b.a charge q of -5/10^6 is sort perpendicular to the line joining a and b thought c with kinetic energy of 0.06j.the charge q comes to rest at a point d.find the distance CD
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"Thought" ??? What it means?
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Answered by
264
see diagram.
I suppose the charge q = - 5 μC starts with a KE of 0.06 J from point C and stops at point d.
Initial total energy of the charge q = KE + PE
= 0.06 + 9*10⁹ * 5 *10⁻⁶ * (-5 *10⁻⁶)/3 + 9*10⁹ * 5*10⁻⁶*(-5*10⁻⁶)/3 J
= 0.06 - 0.075 - 0.075 J
= - 0.09 J
Let ad = bd = x m
Final energy = PE (as KE = 0)
= - 2 * 9*10⁹ * 5*10⁻⁶ * 5 * 10⁻⁶ /x J
= - 0.450/x J
Hence, using Energy conservation we get:
0.450/x = 0.09
x = 5 m
cd = √(x² - 3²) = 4 m
I suppose the charge q = - 5 μC starts with a KE of 0.06 J from point C and stops at point d.
Initial total energy of the charge q = KE + PE
= 0.06 + 9*10⁹ * 5 *10⁻⁶ * (-5 *10⁻⁶)/3 + 9*10⁹ * 5*10⁻⁶*(-5*10⁻⁶)/3 J
= 0.06 - 0.075 - 0.075 J
= - 0.09 J
Let ad = bd = x m
Final energy = PE (as KE = 0)
= - 2 * 9*10⁹ * 5*10⁻⁶ * 5 * 10⁻⁶ /x J
= - 0.450/x J
Hence, using Energy conservation we get:
0.450/x = 0.09
x = 5 m
cd = √(x² - 3²) = 4 m
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clik on red heart thanks above pls
nice answer sir
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Answered by
39
Explanation:
balance energy before and after.
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