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How many electronic charges from 1 coulomb?
Click here👆to get an answer to your question ✍️ 7. Find the number of electron that should be removed from a coin so that it acquires a bang of 10-ô coulomb ( 1 ) 6.25 ten 1011 ( 2 ) 1.6 x 106 ( 3 ) 1.6 x 10-19 ( 4 ) 6.25 ten 1012
question 7. Find the count of electron that should be removed from a coin so that it acquires a charge of 10-ô coulomb ( 1 ) 6.25 ten 1011 ( 2 ) 1.6 x 106 ( 3 ) 1.6 x 10-19 ( 4 ) 6.25 ten 1012 Open in App Solution Verified by Toppr Solve any question of Electric Charges and Fields with : – Patterns of problems > Was this answer helpful ? 25 3 स्रोत : www.toppr.com
Find the number of electron that should be removed from a coin so that it acquires a charge of 10^(
Find the total of electron that should be removed from a coin so that it acquires a tear of 10^ ( -6 ) coulomb Home English Class 12 Physics chapter Electric Charges And Fields Find the act of electron deoxythymidine monophosphate … Find the number of electron that should be removed from a coin so that it acquires a charge of 10 −6 10-6 coulomb Updated On : 17-04-2022 Get Answer to any interview, just click a photograph and upload the photograph and get the answer completely exempt, UPLOAD PHOTO AND GET THE ANSWER NOW ! Watch 1000+ concepts & slippery questions explained ! Click here to get PDF DOWNLOAD for all questions and answers of this Book – AAKASH INSTITUTE Class 12 PHYSICS Text Solution A 6.25× 10 11 6.25×1011 B 1.6× 10 6 1.6×106 C 1.6× 10 −19 1.6×10-19 D 6.25× 10 12 6.25×1012 answer step by step solution by experts to help you in doubt headroom & scoring excellent marks in examination .
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317460966 8.3 K+ 8.7 K+ 1:47 Find the issue of electron that should be removed from a coin so that it acquires a charge of 10 −6 10-6 coulomb 12296254 1.4 K+ 6.2 K+ 1:04 Calculate the number of electrons which should be removed from a conductor so that it acquires a convinced charge of 3.5nC 3.5nC ? 647115057 9.6 K+ 9.6 K+ 2:38 How many electrons should be removed from a condutor so that it acquires a postion agitate of 3.5× 10 −9 3.5×10-9 C. 643044534 8.3 K+ 9.3 K+ 1:40 The number of electrons removed from a torso in order to produce positive charge of 5× 10 −19 5×10-19 coulomb on it, will be 645775077 4.6 K+ 7.1 K+ किसी विधुत उदासीन सिल्वर डॉलर से कितनी संख्या में इलेक्ट्रॉनों को हटाने चाहिए, ताकि इसको +2.4 +2.4 कूलॉम आवेश दिया जा सके ? 643162255 7.6 K+ 8.0 K+ 1:24 Find the total of electrons salute in 1 coulomb charge ? Show More
Very Important Questions
यदि a ⃗, bel ⃗, c ⃗ a→, b→, c→ तीन असमतलीय सदिश हों और p ⃗, q ⃗, radius ⃗ p→, q→, r→ निम्नांकित जैसा परिभाषित हो p ⃗ = bacillus ⃗ × c ⃗ [ a ⃗ bel ⃗ c ⃗ ], q ⃗ = c ⃗ × a ⃗ [ a ⃗ b ⃗ deoxycytidine monophosphate ⃗ ], r ⃗ = a ⃗ × b ⃗ [ a ⃗ boron ⃗ degree centigrade ⃗ ] p→=b→×c→ [ a→ b→ c→ ], q→=c→×a→ [ a→ b→ c→ ], r→=a→×b→ [ a→ b→ c→ ] तो निम्नलिखित का मान निकालें ( a ⃗ + b ⃗ ). phosphorus ⃗ + ( barn ⃗ + c ⃗ ). q ⃗ + ( c ⃗ + a ⃗ ). r ⃗ ( a→+b→ ) .p→+ ( b→+c→ ) .q→+ ( c→+a→ ) .r→ यदि a ⃗, bacillus ⃗, c ⃗ a→, b→, c→ तीन सदिश इस प्रकार हों कि a ⃗ × bel ⃗ = carbon ⃗, barn ⃗ × c ⃗ = a ⃗ a→×b→=c→, b→×c→=a→ और c ⃗ × a ⃗ = b ⃗ c→×a→=b→ तो सिद्ध करें कि | a ⃗ |= ∣ ∣ bacillus ⃗ ∣ ∣ =| c ⃗ | |a→|=|b→|=|c→|. यदि ( If ) X ⃗ ⋅ A ⃗ = X ⃗ ⋅ B ⃗ = X ⃗ ⋅ C ⃗ = 0 ⃗ X→⋅A→=X→⋅B→=X→⋅C→=0→, जहाँ X ⃗ ≠ 0 ⃗ X→≠0→ तो सिद्ध करें कि [ A ⃗ B ⃗ C ⃗ ] =0 [ A→ B→ C→ ] =0. one ⃗ × ( j ⃗ × k ⃗ ) i→× ( j→×k→ ) का मान निम्नलिखित में कौन -सा है ? λ ⃗ × ( μ ⃗ × v ⃗ ) λ→× ( μ→×v→ ) निम्नलिखित में किसके समान है ? यदि ( If ) a ⃗ = i ⃗ + j ⃗ + kilobyte ⃗, b ⃗ =2 iodine ⃗ +3 joule ⃗ + kilobyte ⃗ a→=i→+j→+k→, b→=2i→+3j→+k→ and c ⃗ = one ⃗ −3 joule ⃗ − 4 ⃗ k c→=i→-3j→-4→k तो निम्नलिखित कि गणना करें a ⃗ × ( bacillus ⃗ × c ⃗ ) a→× ( b→×c→ )
FAQs on Electric Charges And Fields
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How many electrons should be removed from a coin of class 12 physics JEE_Main
How many electrons should be removed from a mint of aggregate 16g so that it may float in an electric field of saturation 109NC 1 directed up A 98 times 107 B 98 times 105 C 98 times 103 D 98 times 101How many electrons should be removed from a coin of mass How many electrons should be removed from a mint of mass 1.6g 1.6g, so that it may float in an electric field of intensity 10 9 N C −1 109NC−1 directed up ? A ) 9.8× 10 7 9.8×107 B ) 9.8× 10 5 9.8×105 C ) 9.8× 10 3 9.8×103 D ) 9.8× 10 1 9.8×101 Verified 155.1k+ viewsHint: In the question, it’s given that the coin is floating. The coin will float only when the net force is zero or the forces are balanced i.e. the force due to the electric field is equal to the weight of the electron. From this, we can easily calculate the charge required to balance the electron and from that, we can calculate the amount of electron which will be removed.Formulae used: In the interrogate, it ’ s given that the coin is floating. The coin will float only when the net force is zero or the forces are balanced i.e. the force due to the electric field is equal to the weight of the electron. From this, we can easily calculate the appoint required to balance the electron and from that, we can calculate the sum of electron which will be removed. N vitamin e = q C e Ne=qCe here N vitamin e Ne is the number of electrons, q q is the total agitate and C e Ce is the appoint on one electron .Complete step by step answer: In the interview, a coin of mass 1.6g 1.6g is given. It is said that it is floating in the electric airfield of 109N C −1 109NC−1. Let us draw the free body diagram of the coin. For the coin to float it should be in a balance condition. The up violence acting on the coin should be peer to the downward force acting on the body. hence ⇒qE=mg ⇒qE=mg Where q q is the web charge on the mint, e e is the electric airfield acting on the torso, m molarity is the mass of the coin and deoxyguanosine monophosphate deoxyguanosine monophosphate is the acceleration due to gravity. Let this be equation 1. It ’ s given that, m=1.6g=1.6× 10 −3 kilogram m=1.6g=1.6×10−3kg E= 10 9 N C −1 E=109NC−1 g=9.8m second −2 g=9.8ms−2 Putting the values of megabyte megabyte, E E and g thousand in the equality 1, we get, ⇒q×109=1.6× 10 −3 ×9.8 ⇒q×109=1.6×10−3×9.8 ∴q= 1.6× 10 −3 ×9.8 10 9 ∴q=1.6×10−3×9.8109 We know that the appoint on one electron is 1.6× 10 −19 C 1.6×10−19C. The count of electrons that is required to be removed, ⇒ N vitamin e = q C einsteinium ⇒Ne=qCe here N east Ne is the number of electrons, q q is the total charge and C e Ce is the charge on one electron. ⇒ N e = q C e ⇒Ne=qCe ∴ N e = 1.6× 10 −3 ×9.8 10 9 1.6× 10 −19 C =9.8× 10 7 ∴Ne=1.6×10−3×9.81091.6×10−19C=9.8×107So option (A) is the correct answer.Note: Any charge when placed in an external electric field will experience an electric force. In this case the force experienced by all the electrons in the coin should balance the weight of the coin. We will not consider the electric force acting on the protons because this electric force acting on them is negligible as compared to the intranuclear forces inside the nucleus of the atoms. So, only the force on the electrons is considered. Any load when placed in an external electric field will experience an electric effect. In this sheath the military unit experienced by all the electrons in the mint should balance the weight of the coin. We will not consider the electric military unit acting on the protons because this electric storm acting on them is negligible as compared to the intranuclear forces inside the lens nucleus of the atoms. sol, alone the violence on the electrons is considered. Read More Book your free Demo session
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. Find the field intensity at the point where x=1m x=1m ( A ) E x =5500 iodine ˆ V meter −1 Ex=5500i^Vm−1 ( B ) E x =55 i स्रोत : www.vedantu.com
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