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Q.A point charge of 3 x 10^-9 C is released from rest in a uniform electric field and moves a distance of 5 cm after which its kinetic energy becomes 4.5 x 10^-5 J. Calculate the magnitude of the electric field.

Odisha ChseOdisha CHSE +2 Science Board Exam 2018Subjective· 2mImportance★★★★★
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Work-energy theorem: qEd = KE, so E = KE/(qd) = 3 x 10^5 N/C.

When a charge starts from rest and moves through distance dd in a uniform field EE, the constant electric force qEqE does work that becomes kinetic energy:

W=qEd=ΔKEW = qEd = \Delta KE

Step 1 — Solve for the field: E=ΔKEq dE = \dfrac{\Delta KE}{q\,d}.

Step 2 — Substitute ΔKE=4.5×10−5 J\Delta KE = 4.5\times10^{-5}\ \text{J}, q=3×10−9 Cq = 3\times10^{-9}\ \text{C}, d=5 cm=0.05 md = 5\ \text{cm} = 0.05\ \text{m}:

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