358002
Two conducting spheres of radii \({r_1}\,{\rm{and}}\,{r_2}\) are charged to the same surface charge density. The ratio of electric fields near their surface is
358003
A charge \(Q\) is uniformly distributed in a hollow sphere of radii \({r_1}\) and \({r_2}({r_2} > {r_1})\). The electric field at a point \(P\) distance \(x\) from the centre for \({r_1} < x < {r_2}\) is
358004
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of \(25.5\,KV{m^{ - 1}}\). The density of liquid is \(1.26 \times {10^3}kg{m^{ - 3}}\).The radius of the drop is (neglect buoyancy).
1 \(4.3 \times {10^{ - 7}}\,\,m\)
2 \(7.8 \times {10^{ - 7}}\,\,m\)
3 \(0.078 \times {10^{ - 7}}\,\,m\)
4 \(3.4 \times {10^{ - 7}}\,\,m\)
Explanation:
\(F = qE = mg\,(q = 6e = 6 \times 1.6 \times {10^{ - 19}})\) \(Density(d)\, = \,\frac{{mass}}{{volume}} = \frac{m}{{\frac{4}{3}\pi {r^3}}}or\,{r^3} = \,\frac{m}{{\frac{4}{3}\pi d}}\) Putting the value of \(d\) and \(m\) \(\left( { = \frac{{qE}}{g}} \right)\) and solving we get \(r = 7.8 \times {10^{ - 7}}m\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358005
In a region with uniform electric field, the number of lines of force per unit area is \(E\). If a spherical metallic conductor is placed in this region, the number of lines of force per unit area inside the conductor will be
1 More than \(E\)
2 Zero
3 \(E\)
4 Less than \(E\)
Explanation:
Electric field inside the metallic conductor is zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358006
A positively charged ball hangs from a silk thread. We put a positive test charge \(q_{0}\) at a point and measure \(F / q_{0}\), then it can be predicted that the electric field strength \(E\)
1 \(>F / q_{0}\)
2 \(=\dfrac{F}{q}\)
3 \( < F / q_{0}\)
4 Cannot be estimated
Explanation:
Due to presence of test charge \(q_{0}\) in front of positively charged ball, there would be a redistribution of charge on the ball. In the redistribution of charge, there will be less charge on front half surface and more charge on the back half surface. As a result, the net force \(F\) between ball and charge will decrease, \(i.e.\) the electric field is decreased. Thus, actual electric field will be greater than \(F/{q_0}\)
358002
Two conducting spheres of radii \({r_1}\,{\rm{and}}\,{r_2}\) are charged to the same surface charge density. The ratio of electric fields near their surface is
358003
A charge \(Q\) is uniformly distributed in a hollow sphere of radii \({r_1}\) and \({r_2}({r_2} > {r_1})\). The electric field at a point \(P\) distance \(x\) from the centre for \({r_1} < x < {r_2}\) is
358004
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of \(25.5\,KV{m^{ - 1}}\). The density of liquid is \(1.26 \times {10^3}kg{m^{ - 3}}\).The radius of the drop is (neglect buoyancy).
1 \(4.3 \times {10^{ - 7}}\,\,m\)
2 \(7.8 \times {10^{ - 7}}\,\,m\)
3 \(0.078 \times {10^{ - 7}}\,\,m\)
4 \(3.4 \times {10^{ - 7}}\,\,m\)
Explanation:
\(F = qE = mg\,(q = 6e = 6 \times 1.6 \times {10^{ - 19}})\) \(Density(d)\, = \,\frac{{mass}}{{volume}} = \frac{m}{{\frac{4}{3}\pi {r^3}}}or\,{r^3} = \,\frac{m}{{\frac{4}{3}\pi d}}\) Putting the value of \(d\) and \(m\) \(\left( { = \frac{{qE}}{g}} \right)\) and solving we get \(r = 7.8 \times {10^{ - 7}}m\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358005
In a region with uniform electric field, the number of lines of force per unit area is \(E\). If a spherical metallic conductor is placed in this region, the number of lines of force per unit area inside the conductor will be
1 More than \(E\)
2 Zero
3 \(E\)
4 Less than \(E\)
Explanation:
Electric field inside the metallic conductor is zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358006
A positively charged ball hangs from a silk thread. We put a positive test charge \(q_{0}\) at a point and measure \(F / q_{0}\), then it can be predicted that the electric field strength \(E\)
1 \(>F / q_{0}\)
2 \(=\dfrac{F}{q}\)
3 \( < F / q_{0}\)
4 Cannot be estimated
Explanation:
Due to presence of test charge \(q_{0}\) in front of positively charged ball, there would be a redistribution of charge on the ball. In the redistribution of charge, there will be less charge on front half surface and more charge on the back half surface. As a result, the net force \(F\) between ball and charge will decrease, \(i.e.\) the electric field is decreased. Thus, actual electric field will be greater than \(F/{q_0}\)
358002
Two conducting spheres of radii \({r_1}\,{\rm{and}}\,{r_2}\) are charged to the same surface charge density. The ratio of electric fields near their surface is
358003
A charge \(Q\) is uniformly distributed in a hollow sphere of radii \({r_1}\) and \({r_2}({r_2} > {r_1})\). The electric field at a point \(P\) distance \(x\) from the centre for \({r_1} < x < {r_2}\) is
358004
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of \(25.5\,KV{m^{ - 1}}\). The density of liquid is \(1.26 \times {10^3}kg{m^{ - 3}}\).The radius of the drop is (neglect buoyancy).
1 \(4.3 \times {10^{ - 7}}\,\,m\)
2 \(7.8 \times {10^{ - 7}}\,\,m\)
3 \(0.078 \times {10^{ - 7}}\,\,m\)
4 \(3.4 \times {10^{ - 7}}\,\,m\)
Explanation:
\(F = qE = mg\,(q = 6e = 6 \times 1.6 \times {10^{ - 19}})\) \(Density(d)\, = \,\frac{{mass}}{{volume}} = \frac{m}{{\frac{4}{3}\pi {r^3}}}or\,{r^3} = \,\frac{m}{{\frac{4}{3}\pi d}}\) Putting the value of \(d\) and \(m\) \(\left( { = \frac{{qE}}{g}} \right)\) and solving we get \(r = 7.8 \times {10^{ - 7}}m\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358005
In a region with uniform electric field, the number of lines of force per unit area is \(E\). If a spherical metallic conductor is placed in this region, the number of lines of force per unit area inside the conductor will be
1 More than \(E\)
2 Zero
3 \(E\)
4 Less than \(E\)
Explanation:
Electric field inside the metallic conductor is zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358006
A positively charged ball hangs from a silk thread. We put a positive test charge \(q_{0}\) at a point and measure \(F / q_{0}\), then it can be predicted that the electric field strength \(E\)
1 \(>F / q_{0}\)
2 \(=\dfrac{F}{q}\)
3 \( < F / q_{0}\)
4 Cannot be estimated
Explanation:
Due to presence of test charge \(q_{0}\) in front of positively charged ball, there would be a redistribution of charge on the ball. In the redistribution of charge, there will be less charge on front half surface and more charge on the back half surface. As a result, the net force \(F\) between ball and charge will decrease, \(i.e.\) the electric field is decreased. Thus, actual electric field will be greater than \(F/{q_0}\)
358002
Two conducting spheres of radii \({r_1}\,{\rm{and}}\,{r_2}\) are charged to the same surface charge density. The ratio of electric fields near their surface is
358003
A charge \(Q\) is uniformly distributed in a hollow sphere of radii \({r_1}\) and \({r_2}({r_2} > {r_1})\). The electric field at a point \(P\) distance \(x\) from the centre for \({r_1} < x < {r_2}\) is
358004
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of \(25.5\,KV{m^{ - 1}}\). The density of liquid is \(1.26 \times {10^3}kg{m^{ - 3}}\).The radius of the drop is (neglect buoyancy).
1 \(4.3 \times {10^{ - 7}}\,\,m\)
2 \(7.8 \times {10^{ - 7}}\,\,m\)
3 \(0.078 \times {10^{ - 7}}\,\,m\)
4 \(3.4 \times {10^{ - 7}}\,\,m\)
Explanation:
\(F = qE = mg\,(q = 6e = 6 \times 1.6 \times {10^{ - 19}})\) \(Density(d)\, = \,\frac{{mass}}{{volume}} = \frac{m}{{\frac{4}{3}\pi {r^3}}}or\,{r^3} = \,\frac{m}{{\frac{4}{3}\pi d}}\) Putting the value of \(d\) and \(m\) \(\left( { = \frac{{qE}}{g}} \right)\) and solving we get \(r = 7.8 \times {10^{ - 7}}m\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358005
In a region with uniform electric field, the number of lines of force per unit area is \(E\). If a spherical metallic conductor is placed in this region, the number of lines of force per unit area inside the conductor will be
1 More than \(E\)
2 Zero
3 \(E\)
4 Less than \(E\)
Explanation:
Electric field inside the metallic conductor is zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358006
A positively charged ball hangs from a silk thread. We put a positive test charge \(q_{0}\) at a point and measure \(F / q_{0}\), then it can be predicted that the electric field strength \(E\)
1 \(>F / q_{0}\)
2 \(=\dfrac{F}{q}\)
3 \( < F / q_{0}\)
4 Cannot be estimated
Explanation:
Due to presence of test charge \(q_{0}\) in front of positively charged ball, there would be a redistribution of charge on the ball. In the redistribution of charge, there will be less charge on front half surface and more charge on the back half surface. As a result, the net force \(F\) between ball and charge will decrease, \(i.e.\) the electric field is decreased. Thus, actual electric field will be greater than \(F/{q_0}\)
358002
Two conducting spheres of radii \({r_1}\,{\rm{and}}\,{r_2}\) are charged to the same surface charge density. The ratio of electric fields near their surface is
358003
A charge \(Q\) is uniformly distributed in a hollow sphere of radii \({r_1}\) and \({r_2}({r_2} > {r_1})\). The electric field at a point \(P\) distance \(x\) from the centre for \({r_1} < x < {r_2}\) is
358004
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of \(25.5\,KV{m^{ - 1}}\). The density of liquid is \(1.26 \times {10^3}kg{m^{ - 3}}\).The radius of the drop is (neglect buoyancy).
1 \(4.3 \times {10^{ - 7}}\,\,m\)
2 \(7.8 \times {10^{ - 7}}\,\,m\)
3 \(0.078 \times {10^{ - 7}}\,\,m\)
4 \(3.4 \times {10^{ - 7}}\,\,m\)
Explanation:
\(F = qE = mg\,(q = 6e = 6 \times 1.6 \times {10^{ - 19}})\) \(Density(d)\, = \,\frac{{mass}}{{volume}} = \frac{m}{{\frac{4}{3}\pi {r^3}}}or\,{r^3} = \,\frac{m}{{\frac{4}{3}\pi d}}\) Putting the value of \(d\) and \(m\) \(\left( { = \frac{{qE}}{g}} \right)\) and solving we get \(r = 7.8 \times {10^{ - 7}}m\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358005
In a region with uniform electric field, the number of lines of force per unit area is \(E\). If a spherical metallic conductor is placed in this region, the number of lines of force per unit area inside the conductor will be
1 More than \(E\)
2 Zero
3 \(E\)
4 Less than \(E\)
Explanation:
Electric field inside the metallic conductor is zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358006
A positively charged ball hangs from a silk thread. We put a positive test charge \(q_{0}\) at a point and measure \(F / q_{0}\), then it can be predicted that the electric field strength \(E\)
1 \(>F / q_{0}\)
2 \(=\dfrac{F}{q}\)
3 \( < F / q_{0}\)
4 Cannot be estimated
Explanation:
Due to presence of test charge \(q_{0}\) in front of positively charged ball, there would be a redistribution of charge on the ball. In the redistribution of charge, there will be less charge on front half surface and more charge on the back half surface. As a result, the net force \(F\) between ball and charge will decrease, \(i.e.\) the electric field is decreased. Thus, actual electric field will be greater than \(F/{q_0}\)
