\(\begin{aligned}F_{n e t} & =2 F \cos \theta=2 \times \dfrac{K Q q}{\left(a^{2}+x^{2}\right)} \dfrac{x}{\sqrt{a^{2}+x^{2}}} \\& =\dfrac{2 K Q q x}{\left(a^{2}+x^{2}\right)^{3 / 2}}\end{aligned}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358083
Two point charges \( + 3\,\mu C\) and \( + 8\,\mu C\) repel each other with a force of 40 \(N\) . If a charge of \( - 5\,\mu C\) is added to each of them, then the force between them will become
1 \( - 10\,N\)
2 \( + 10\,N\)
3 \( + 20\,N\)
4 \( - 20\,N\)
Explanation:
\({ F=\dfrac{k q_{1} q_{2}}{r^{2}}}\)
\({40=k \dfrac{q_{1} q_{2}}{r^{2}} \Rightarrow r^{2}=\dfrac{k(3 \times 8)}{40}}\) If \({-5 C}\) is added to each charge \({F^{\prime}=k \dfrac{\left[q_{1}+(-5)\right]\left[q_{2}+(-5)\right]}{r^{2}}}\) \({=k \dfrac{[3-5][8-5]}{r^{2}}=\dfrac{k(-2) \times(3)}{k \times 3 \times 8} \times 40}\) \({F^{\prime}=-10 {~N}}\). So correct option is (1)
PHXII01:ELECTRIC CHARGES AND FIELDS
358084
The ratio of electric force between two electrons to two protons separated by the same distance in air is
1 \({10^0}\)
2 \({10^6}\)
3 \({10^4}\)
4 None of these
Explanation:
As charges on both electrons and protons are same so the forces are also same.
PHXII01:ELECTRIC CHARGES AND FIELDS
358085
Assertion : If a conducting medium is placed between two charges, then electric force between them becomes zero. Reason : Reduction in a force due to introduced material is inversely proportional to its dielectric constant. conductor is not dielectric.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Such metals, \(\epsilon_{r}=\infty\) \(F^{\prime}=\left(F / \epsilon_{r}\right) \rightarrow 0\) But this is due to \(F^{\prime}=\dfrac{1}{\epsilon_{r}}\) So correct option is (1).
PHXII01:ELECTRIC CHARGES AND FIELDS
358086
The nucleus of helium atom contains two protons that are separated by distance \(3.0 \times {10^{ - 15}}m\). The magnitude of the electrostatic force that each proton exerts on the other is
1 \(25.6\,N\)
2 \(20.6\,N\)
3 \(12.6\,N\)
4 \(15.6\,N\)
Explanation:
Charge of proton, \({q_p} = 1.6 \times {10^{ - 19}}\,C\) Distance between the protons, \(r = 3 \times {10^{ - 15}}\,m\) The magnitude of electrostatic force between then is \({{\rm{F}}_e} = \frac{{{q_p}{q_p}}}{{4\pi {\varepsilon _0}{r^2}}} = \frac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{{{\left( {3 \times {{10}^{ - 15}}} \right)}^2}}}\) \( = 25.6\,N\)
\(\begin{aligned}F_{n e t} & =2 F \cos \theta=2 \times \dfrac{K Q q}{\left(a^{2}+x^{2}\right)} \dfrac{x}{\sqrt{a^{2}+x^{2}}} \\& =\dfrac{2 K Q q x}{\left(a^{2}+x^{2}\right)^{3 / 2}}\end{aligned}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358083
Two point charges \( + 3\,\mu C\) and \( + 8\,\mu C\) repel each other with a force of 40 \(N\) . If a charge of \( - 5\,\mu C\) is added to each of them, then the force between them will become
1 \( - 10\,N\)
2 \( + 10\,N\)
3 \( + 20\,N\)
4 \( - 20\,N\)
Explanation:
\({ F=\dfrac{k q_{1} q_{2}}{r^{2}}}\)
\({40=k \dfrac{q_{1} q_{2}}{r^{2}} \Rightarrow r^{2}=\dfrac{k(3 \times 8)}{40}}\) If \({-5 C}\) is added to each charge \({F^{\prime}=k \dfrac{\left[q_{1}+(-5)\right]\left[q_{2}+(-5)\right]}{r^{2}}}\) \({=k \dfrac{[3-5][8-5]}{r^{2}}=\dfrac{k(-2) \times(3)}{k \times 3 \times 8} \times 40}\) \({F^{\prime}=-10 {~N}}\). So correct option is (1)
PHXII01:ELECTRIC CHARGES AND FIELDS
358084
The ratio of electric force between two electrons to two protons separated by the same distance in air is
1 \({10^0}\)
2 \({10^6}\)
3 \({10^4}\)
4 None of these
Explanation:
As charges on both electrons and protons are same so the forces are also same.
PHXII01:ELECTRIC CHARGES AND FIELDS
358085
Assertion : If a conducting medium is placed between two charges, then electric force between them becomes zero. Reason : Reduction in a force due to introduced material is inversely proportional to its dielectric constant. conductor is not dielectric.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Such metals, \(\epsilon_{r}=\infty\) \(F^{\prime}=\left(F / \epsilon_{r}\right) \rightarrow 0\) But this is due to \(F^{\prime}=\dfrac{1}{\epsilon_{r}}\) So correct option is (1).
PHXII01:ELECTRIC CHARGES AND FIELDS
358086
The nucleus of helium atom contains two protons that are separated by distance \(3.0 \times {10^{ - 15}}m\). The magnitude of the electrostatic force that each proton exerts on the other is
1 \(25.6\,N\)
2 \(20.6\,N\)
3 \(12.6\,N\)
4 \(15.6\,N\)
Explanation:
Charge of proton, \({q_p} = 1.6 \times {10^{ - 19}}\,C\) Distance between the protons, \(r = 3 \times {10^{ - 15}}\,m\) The magnitude of electrostatic force between then is \({{\rm{F}}_e} = \frac{{{q_p}{q_p}}}{{4\pi {\varepsilon _0}{r^2}}} = \frac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{{{\left( {3 \times {{10}^{ - 15}}} \right)}^2}}}\) \( = 25.6\,N\)
\(\begin{aligned}F_{n e t} & =2 F \cos \theta=2 \times \dfrac{K Q q}{\left(a^{2}+x^{2}\right)} \dfrac{x}{\sqrt{a^{2}+x^{2}}} \\& =\dfrac{2 K Q q x}{\left(a^{2}+x^{2}\right)^{3 / 2}}\end{aligned}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358083
Two point charges \( + 3\,\mu C\) and \( + 8\,\mu C\) repel each other with a force of 40 \(N\) . If a charge of \( - 5\,\mu C\) is added to each of them, then the force between them will become
1 \( - 10\,N\)
2 \( + 10\,N\)
3 \( + 20\,N\)
4 \( - 20\,N\)
Explanation:
\({ F=\dfrac{k q_{1} q_{2}}{r^{2}}}\)
\({40=k \dfrac{q_{1} q_{2}}{r^{2}} \Rightarrow r^{2}=\dfrac{k(3 \times 8)}{40}}\) If \({-5 C}\) is added to each charge \({F^{\prime}=k \dfrac{\left[q_{1}+(-5)\right]\left[q_{2}+(-5)\right]}{r^{2}}}\) \({=k \dfrac{[3-5][8-5]}{r^{2}}=\dfrac{k(-2) \times(3)}{k \times 3 \times 8} \times 40}\) \({F^{\prime}=-10 {~N}}\). So correct option is (1)
PHXII01:ELECTRIC CHARGES AND FIELDS
358084
The ratio of electric force between two electrons to two protons separated by the same distance in air is
1 \({10^0}\)
2 \({10^6}\)
3 \({10^4}\)
4 None of these
Explanation:
As charges on both electrons and protons are same so the forces are also same.
PHXII01:ELECTRIC CHARGES AND FIELDS
358085
Assertion : If a conducting medium is placed between two charges, then electric force between them becomes zero. Reason : Reduction in a force due to introduced material is inversely proportional to its dielectric constant. conductor is not dielectric.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Such metals, \(\epsilon_{r}=\infty\) \(F^{\prime}=\left(F / \epsilon_{r}\right) \rightarrow 0\) But this is due to \(F^{\prime}=\dfrac{1}{\epsilon_{r}}\) So correct option is (1).
PHXII01:ELECTRIC CHARGES AND FIELDS
358086
The nucleus of helium atom contains two protons that are separated by distance \(3.0 \times {10^{ - 15}}m\). The magnitude of the electrostatic force that each proton exerts on the other is
1 \(25.6\,N\)
2 \(20.6\,N\)
3 \(12.6\,N\)
4 \(15.6\,N\)
Explanation:
Charge of proton, \({q_p} = 1.6 \times {10^{ - 19}}\,C\) Distance between the protons, \(r = 3 \times {10^{ - 15}}\,m\) The magnitude of electrostatic force between then is \({{\rm{F}}_e} = \frac{{{q_p}{q_p}}}{{4\pi {\varepsilon _0}{r^2}}} = \frac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{{{\left( {3 \times {{10}^{ - 15}}} \right)}^2}}}\) \( = 25.6\,N\)
\(\begin{aligned}F_{n e t} & =2 F \cos \theta=2 \times \dfrac{K Q q}{\left(a^{2}+x^{2}\right)} \dfrac{x}{\sqrt{a^{2}+x^{2}}} \\& =\dfrac{2 K Q q x}{\left(a^{2}+x^{2}\right)^{3 / 2}}\end{aligned}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358083
Two point charges \( + 3\,\mu C\) and \( + 8\,\mu C\) repel each other with a force of 40 \(N\) . If a charge of \( - 5\,\mu C\) is added to each of them, then the force between them will become
1 \( - 10\,N\)
2 \( + 10\,N\)
3 \( + 20\,N\)
4 \( - 20\,N\)
Explanation:
\({ F=\dfrac{k q_{1} q_{2}}{r^{2}}}\)
\({40=k \dfrac{q_{1} q_{2}}{r^{2}} \Rightarrow r^{2}=\dfrac{k(3 \times 8)}{40}}\) If \({-5 C}\) is added to each charge \({F^{\prime}=k \dfrac{\left[q_{1}+(-5)\right]\left[q_{2}+(-5)\right]}{r^{2}}}\) \({=k \dfrac{[3-5][8-5]}{r^{2}}=\dfrac{k(-2) \times(3)}{k \times 3 \times 8} \times 40}\) \({F^{\prime}=-10 {~N}}\). So correct option is (1)
PHXII01:ELECTRIC CHARGES AND FIELDS
358084
The ratio of electric force between two electrons to two protons separated by the same distance in air is
1 \({10^0}\)
2 \({10^6}\)
3 \({10^4}\)
4 None of these
Explanation:
As charges on both electrons and protons are same so the forces are also same.
PHXII01:ELECTRIC CHARGES AND FIELDS
358085
Assertion : If a conducting medium is placed between two charges, then electric force between them becomes zero. Reason : Reduction in a force due to introduced material is inversely proportional to its dielectric constant. conductor is not dielectric.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Such metals, \(\epsilon_{r}=\infty\) \(F^{\prime}=\left(F / \epsilon_{r}\right) \rightarrow 0\) But this is due to \(F^{\prime}=\dfrac{1}{\epsilon_{r}}\) So correct option is (1).
PHXII01:ELECTRIC CHARGES AND FIELDS
358086
The nucleus of helium atom contains two protons that are separated by distance \(3.0 \times {10^{ - 15}}m\). The magnitude of the electrostatic force that each proton exerts on the other is
1 \(25.6\,N\)
2 \(20.6\,N\)
3 \(12.6\,N\)
4 \(15.6\,N\)
Explanation:
Charge of proton, \({q_p} = 1.6 \times {10^{ - 19}}\,C\) Distance between the protons, \(r = 3 \times {10^{ - 15}}\,m\) The magnitude of electrostatic force between then is \({{\rm{F}}_e} = \frac{{{q_p}{q_p}}}{{4\pi {\varepsilon _0}{r^2}}} = \frac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{{{\left( {3 \times {{10}^{ - 15}}} \right)}^2}}}\) \( = 25.6\,N\)
\(\begin{aligned}F_{n e t} & =2 F \cos \theta=2 \times \dfrac{K Q q}{\left(a^{2}+x^{2}\right)} \dfrac{x}{\sqrt{a^{2}+x^{2}}} \\& =\dfrac{2 K Q q x}{\left(a^{2}+x^{2}\right)^{3 / 2}}\end{aligned}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358083
Two point charges \( + 3\,\mu C\) and \( + 8\,\mu C\) repel each other with a force of 40 \(N\) . If a charge of \( - 5\,\mu C\) is added to each of them, then the force between them will become
1 \( - 10\,N\)
2 \( + 10\,N\)
3 \( + 20\,N\)
4 \( - 20\,N\)
Explanation:
\({ F=\dfrac{k q_{1} q_{2}}{r^{2}}}\)
\({40=k \dfrac{q_{1} q_{2}}{r^{2}} \Rightarrow r^{2}=\dfrac{k(3 \times 8)}{40}}\) If \({-5 C}\) is added to each charge \({F^{\prime}=k \dfrac{\left[q_{1}+(-5)\right]\left[q_{2}+(-5)\right]}{r^{2}}}\) \({=k \dfrac{[3-5][8-5]}{r^{2}}=\dfrac{k(-2) \times(3)}{k \times 3 \times 8} \times 40}\) \({F^{\prime}=-10 {~N}}\). So correct option is (1)
PHXII01:ELECTRIC CHARGES AND FIELDS
358084
The ratio of electric force between two electrons to two protons separated by the same distance in air is
1 \({10^0}\)
2 \({10^6}\)
3 \({10^4}\)
4 None of these
Explanation:
As charges on both electrons and protons are same so the forces are also same.
PHXII01:ELECTRIC CHARGES AND FIELDS
358085
Assertion : If a conducting medium is placed between two charges, then electric force between them becomes zero. Reason : Reduction in a force due to introduced material is inversely proportional to its dielectric constant. conductor is not dielectric.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Such metals, \(\epsilon_{r}=\infty\) \(F^{\prime}=\left(F / \epsilon_{r}\right) \rightarrow 0\) But this is due to \(F^{\prime}=\dfrac{1}{\epsilon_{r}}\) So correct option is (1).
PHXII01:ELECTRIC CHARGES AND FIELDS
358086
The nucleus of helium atom contains two protons that are separated by distance \(3.0 \times {10^{ - 15}}m\). The magnitude of the electrostatic force that each proton exerts on the other is
1 \(25.6\,N\)
2 \(20.6\,N\)
3 \(12.6\,N\)
4 \(15.6\,N\)
Explanation:
Charge of proton, \({q_p} = 1.6 \times {10^{ - 19}}\,C\) Distance between the protons, \(r = 3 \times {10^{ - 15}}\,m\) The magnitude of electrostatic force between then is \({{\rm{F}}_e} = \frac{{{q_p}{q_p}}}{{4\pi {\varepsilon _0}{r^2}}} = \frac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{{{\left( {3 \times {{10}^{ - 15}}} \right)}^2}}}\) \( = 25.6\,N\)
